Algebra is that branch of mathematics which treats the relation of numbers.
In algebra, two types of symbols are used: constants and variable (literals).
What is Constant?
Constant is a symbol whose value always remains the same, whatever the situation be.
For example: 5, – 9, $latex \frac{3}{8}$, π, , etc.
What is Variable?
Variable is a symbol whose value changes according to the situation.
For example : x, y, z, ax, a + x, 5y, – 7x, etc.
Algebraic Expression
- An algebraic expression is a collection of terms separated by plus (+) or minus (–) sign. For example : 3x + 5y, 7y – 2x, 2x – ay + az, etc.
- The various parts of an algebraic expression that are separated by ‘+’ or ‘–’ sign are called terms.
For example :
Algebraic expression | No. of Terms | Terms |
–32x |
1 | –32x |
2x + 3y | 2 | 2x and 3y |
ax – 5y + cz | 3 | ax, –5y and cz |
$latex \frac{3}{x}+\frac{y}{7}-\frac{{xy}}{8}+9$ | 4 |
$latex \frac{3}{x},\,\frac{y}{7},\,-\frac{{xy}}{8}$ and 9 & so on. |
How Many Types of Algebraic Expressions
Monomial
An algebraic expression having only one term is called a monomial. For ex. 8y, –7xy, 4×2, abx, etc. ‘mono’ means ‘one’.
Binomial
An algebraic expression having two terms is called a Binomial.
For ex.
8x + 3y, 8x + 3, 8 + 3y, a + bz, 9 – 4y,
2×2 – 4z, 6y2 – 5y, etc. ‘bi’ means ‘two’.
Trinomial
An algebraic expression having three terms is called a Trinomial.
For ex.
ax – 5y + 8z, 3×2 + 4x + 7, 9y2 – 3y + 2x, etc.
‘tri means ‘three’.
Multinomial
An algebraic expression having two or more terms is called a Multinomial.
Factor and Coefficient
What is Factor?
Each combination of the constants and variables, which form a term, is called a Factor.
For examples
(i) 7, x and 7x are factors of 7x, in which
7 is constant (numerical) factor and x is variable (literal) factor.
(ii) In –5x2y, the numerical factor is –5 and literal factors are : x, y, xy, x2 and x2y.
² Coefficient :
Any factor of a term is called the coefficient of the remaining term.
For example :
(i) In 7x ; 7 is coefficient of x
(ii) In –5x2y; 5 is coefficient of –x2y; –5 is coefficient of x2y.
Ex. 1 Write the coefficient of :
(i) x2 in 3×3 – 5×2 + 7
(ii) xy in 8xyz
(iii) –y in 2y2 – 6y + 2
(iv) x0 in 3x + 7
Sol. (i) –5
(ii) 8z
(iii) 6
(iv) Since x0 = 1, Therefore
3x + 7 = 3x + 7×0
coefficient of x0 is 7.Ø
The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial.
For example :
(a) In polynomial 5×2 – 8×7 + 3x :
(i) The power of term 5×2 = 2
(ii) The power of term –8×7 = 7
(iii) The power of 3x = 1
Since, the greatest power is 7, therefore degree of the polynomial 5×2 – 8×7 + 3x is 7
(b) The degree of polynomial :
(i) 4y3 – 3y + 8 is 3
(ii) 7p + 2 is 1(p = p1)
(iii) 2m – 7m8 + m13 is 13 and so on.
v EXAMPLES v
Ex.2 Find which of the following algebraic expression is a polynomial.
(i) 3×2 – 5x (ii) x +
(iii) – 8 (iv) z5 – + 8
Sol. (i) 3×2 – 5x = 3×2 – 5×1
It is a polynomial.
(ii) x + = x1 + x–1
It is not a polynomial.
(iii) – 8 = y1/2 – 8
Since, the power of the first term () is , which is not a whole number.
(iv) z5 – + 8 = z5 – z1/3 + 8
Since, the exponent of the second term is
1/3, which in not a whole number. Therefore, the given expression is not a polynomial.
Ex.3 Find the degree of the polynomial :
(i) 5x – 6×3 + 8×7 + 6×2
(ii) 2y12 + 3y10 – y15 + y + 3
(iii) x
(iv) 8
Sol. (i) Since the term with highest exponent (power) is 8×7 and its power is 7.
\ The degree of given polynomial is 7.
(ii) The highest power of the variable is 15
Þ degree = 15.
(iii) x = x1 Þ degree is 1.
(iv) 8 = 8×0 Þ degree = 0
(A) Based on degree :
If degree of polynomial is
Examples | |||
1. | One | Linear | x + 3, y – x + 2, x –3 |
2. | Two | Quadratic | 2x^{2} –7,x^{2}+y^{2} –2xy, x^{2} +1+ 3y |
3. |
Three |
Cubic |
x^{3} + 3x^{2} –7x+8, 2x^{2}+5x^{3}+7, |
4. |
Four |
bi-quadratic |
x^{4} + y^{4} + 2x^{2}y^{2}, x^{4} + 3,… |
(B) Based on Terms :
If number of terms in polynomial is
Examples | |||
1. | One | Monomial | 7x, 5x^{9}, x^{16}, xy, …… |
2. |
Two |
Binomial |
2 + 7y^{6}, y^{3} + x^{14}, 7 + 5x^{9},… |
3. |
Three |
Trinomial |
x^{3} –2x + y, x^{31}+y^{32}+ z^{33},….. |
Note : (1) Degree of constant polynomials
(Ex.5, 7, –3, 8/5, …) is zero.
(2) Degree of zero polynomial (zero = 0
= zero polynomial) is not defined.
If a polynomial has only one variable then it is called polynomial in one variable.
Ex. P(x) = 2x^{3} + 5x – 3 Cubic trinomial
Q(x) = 7x^{7} – 5x^{5} – 3x^{3} + x + 3 polynomial of
degree 7
R(y) = y Linear, monomial
S(t) = t^{2} + 3 Quadratic Binomial
Note : General form of a polynomial in one variable x of degree ‘n’ is a_{n}x^{n} + a_{n–1}x^{n–1 }+ a_{n–2}x^{n–2} + ….+ a_{2}x^{2} + a_{1}x + a_{0}, a_{n} ¹ 0, where a_{n}, a_{n–1},… a_{2}, a_{1}, a_{0} all are constants.
\ for linear ax + b, a ¹ 0
for quadratic ax^{2} + bx + c, a ¹ 0
for cubic ax^{3} + bx^{2} + cx + d, a ¹ 0
Ø
(i) Remainder obtained on dividing polynomial p(x) by x – a is equal to p(a) .
(ii) If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a.
(iii) (x – a) is a factor of polynomial p(x) if p(a) = 0
(iv) (x + a) is a factor of polynomial p(x) if p(–a) = 0
(v) (x – a)(x – b) is a factor of polynomial p(x),
if p(a) = 0 and p(b) = 0.
v EXAMPLES v
Ex.4 Find the remainder when 4×3 – 3×2 + 2x – 4 is divided by
(a) x – 1 (b) x + 2 (c) x +
Sol. Let p(x) = 4×3 – 3×2 + 2x – 4
(a) When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1)
p(1) = 4 (1)3 – 3(1)2 + 2(1) – 4
= 4 × 1 – 3 × 1 + 2 × 1 – 4
= 4 – 3 + 2 – 4 = – 1
(b) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (–2).
p(–2) = 4 (–2)3 – 3 (–2)2 + 2(–2) – 4
= 4 × (–8) – 3 × 4 – 4 – 4
= – 32 – 12 – 8 = – 52
(c) When p(x) is divided by, then by remainder theorem, the required remainder will be
p = 4 – 3 + 2 – 4
= 4 × – 3 × – 2 ×– 4
= –– – 1– 4 = – – 5
= =
Ø For a polynomial f(x) = 3×2 – 4x + 2.
To find its value at x = 3;
replace x by 3 everywhere.
So, the value of f(x) = 3×2 – 4x + 2 at x = 3 is
f(3) = 3 × 32 – 4 × 3 + 2
= 27 – 12 + 2 = 17.
Similarly, the value of polynomial
f(x) = 3×2 – 4x + 2,
(i) at x = –2 is f(–2) = 3(–2)2 –4(–2) + 2
= 12 + 8 + 2 = 22
(ii) at x = 0 is f(0) = 3(0)2 – 4(0) + 2
= 0 – 0 + 2 = 2
(iii) at x = is f = 3– 4 + 2
= – 2 + 2 =
Ex.5 Find the value of the polynomial 5x – 4×2 + 3 at:
(i) x = 0 (ii) x = –1
Sol. Let p(x) = 5x – 4×2 + 3.
(i) At x = 0, p(0) = 5 × 0 – 4 × (0)2 + 3
= 0 – 0 + 3 = 3
(ii) At x = –1, p(–1) = 5(–1) – 4(–1)2 + 3
= –5 – 4 + 3 = – 6
Ø If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x).
For example :
(i) For polynomial p(x) = x – 2; p(2) = 2 – 2 = 0
\ x = 2 or simply 2 is a zero of the polynomial
p(x) = x – 2.
(ii) For the polynomial g(u) = u2 – 5u + 6;
g(3) = (3)2 – 5 × 3 + 6 = 9 – 15 + 6 = 0
\ 3 is a zero of the polynomial g(u)
= u2 – 5u + 6.
Also, g(2) = (2)2 – 5 × 2 + 6 = 4 – 10 + 6 = 0
\ 2 is also a zero of the polynomial
g(u) = u2 – 5u + 6
(a) Every linear polynomial has one and only one zero.
(b) A given polynomial may have more than one zeroes.
(c) If the degree of a polynomial is n; the largest number of zeroes it can have is also n.
For example :
If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a polynomial is 8; largest number of zeroes it can have is 8.
(d) A zero of a polynomial need not be 0.
For example : If f(x) = x2 – 4,
then f(2) = (2)2 – 4 = 4 – 4 = 0
Here, zero of the polynomial f(x) = x2 – 4 is 2 which itself is not 0.
(e) 0 may be a zero of a polynomial.
For example : If f(x) = x2 – x,
then f(0) = 02 – 0 = 0
Here 0 is the zero of polynomial
f(x) = x2 – x.
v EXAMPLES v
Ex.6 Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :
(i) p(x) = 3x + 1, x = –
(ii) p(x) = (x + 1) (x – 2), x = – 1, 2
(iii) p(x) = x2, x = 0
(iv) p(x) = lx + m, x = –
(v) p(x) = 2x + 1, x =
Sol. (i) p(x) = 3x + 1
Þ p = 3 × – + 1 = –1 + 1 = 0
\ x = – is a zero of p(x) = 3x + 1.
(ii) p(x) = (x + 1) (x – 2)
Þ p(–1) = (–1 + 1) (–1 – 2) = 0 × –3 = 0
and, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0
\ x = –1 and x = 2 are zeroes of the given polynomial.
(iii) p(x) = x2 Þ p(0) = 02 = 0
\ x = 0 is a zero of the given polynomial
(iv) p(x) = lx + m Þ p = l + m
= – m + m = 0
\ x = – is a zero of the given polynomial.
(v) p(x) = 2x + 1 Þ p = 2 × + 1
= 1 + 1 = 2 ¹ 0
\ x = is not a zero of the given polynomial.
Ex.7 Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5 (ii) p(x) = 2x + 5
(iii) p(x) = 3x – 2
Sol. To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.
(i) For the zero of polynomial p(x) = x + 5
p(x) = 0 Þ x + 5 = 0 Þ x = –5
\ x = –5 is a zero of the polynomial
p(x) = x + 5.
(ii) p(x) = 0 Þ 2x + 5 = 0
Þ 2x = –5 and x =
\ x = is a zero of p(x) = 2x + 5.
(iii) p(x) = 0 Þ 3x – 2 = 0
Þ 3x = 2 and x = .
\ x = is zero of p(x) = 3x – 2
Let us consider linear polynomial ax + b. The graph of y = ax + b is a straight line.
For example : The graph of y = 3x + 4 is a straight line passing through (0, 4) and (2, 10).
(i) Let us consider the graph of y = 2x – 4 intersects the x-axis at x = 2. The zero 2x – 4 is 2. Thus, the zero of the polynomial 2x – 4 is the x-coordinate of the point where the graph y = 2x – 4 intersects the x-axis.
(ii) A general equation of a linear polynomial is
ax + b. The graph of y = ax + b is a straight line which intersects the x-axis at .
Zero of the polynomial ax + b is the x-coordinate of the point of intersection of the graph with x-axis.
(iii) Let us consider the quadratic polynomial
x2 – 4x + 3. The graph of x2 – 4x + 3 intersects the x-axis at the point (1, 0) and (3, 0). Zeroes of the polynomial x2 – 4x + 3 are the
x-coordinates of the points of intersection of the graph with x-axis.
The shape of the graph of the quadratic polynomials is È and the curve is known as parabola.
(iv) Now let us consider one more polynomial
–x2 + 2x + 8. Graph of this polynomial intersects the x-axis at the points
(4, 0), (–2, 0). Zeroes of the polynomial –x2 + 2x + 8 are the x-coordinates of the points at which the graph intersects the x-axis. The shape of the graph of the given quadratic polynomial is Ç and the curve is known as parabola.
The zeroes of a quadratic polynomial
ax^{2} + bx + c he x-coordinates of the points where the graph of y = ax^{2} + bx + c intersects the x-axis.
Cubic polynomial : Let us find out geometrically how many zeroes a cubic has.
Let consider cubic polynomial
x3 – 6×2 + 11x – 6.
Case 1 :
The graph of the cubic equation intersects the
x-axis at three points (1, 0), (2, 0) and (3, 0). Zeroes of the given polynomial are the
x-coordinates of the points of intersection with the x-axis.
Case 2 :
The cubic equation x3 – x2 intersects the x-axis at the point (0, 0) and (1, 0). Zero of a polynomial x3 – x2 are the x-coordinates of the point where the graph cuts the x-axis.
Zeroes of the cubic polynomial are 0 and 1.
Case 3 :
y = x3
Cubic polynomial has only one zero.
In brief : A cubic equation can have 1 or 2 or 3 zeroes or any polynomial of degree three can have at most three zeroes.
Remarks : In general, polynomial of degree n, the graph of y = p(x) passes x-axis at most at n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.
v EXAMPLES v
Ex.8 Which of the following correspond to the graph to a linear or a quadratic polynomial and find the number of zeroes of polynomial.
(i) (ii)
(iii) (iv)
(v) (vi)
(vii) (viii)
(ix) (x)
Sol. (i) The graph is a straight line so the graph is of a linear polynomial. The number of zeroes is one as the graph intersects the x-axis at one point only.
(ii) The graph is a parabola. So, this is the graph of quadratic polynomial. The number of zeroes is zero as the graph does not intersect the x-axis.
(iii) Here the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points).
(iv) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is two as the graph intersects the x-axis at two points.
(v) The polynomial is linear as the graph is straight line. The number of zeroes is zero as the graph does not intersect the x-axis.
(vi) The polynomial is quadratic as the graph is a parabola. The number of zeroes is 1 as the graph intersects the x-axis at one point (two coincident points) only.
(vii)The polynomial is quadratic as the graph is a parabola. The number of zeroes is zero, as the graph does not intersect the x-axis.
(viii) Polynomial is neither linear nor quadratic as the graph is neither a straight line nor a parabola is one as the graph intersects the x-axis at one point only.
(ix) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points).
(x) The polynomial is linear as the graph is a straight line. The number of zeroes is one as the graph intersects the x-axis at only one point.
Consider quadratic polynomial
P(x) = 2×2 – 16x + 30.
Now, 2×2 – 16x + 30 = (2x – 6) (x – 3)
= 2 (x – 3) (x – 5)
The zeroes of P(x) are 3 and 5.
Sum of the zeroes
= 3 + 5 = 8 = = –
Product of the zeroes
= 3 × 5 = 15 = =
So if ax^{2} + bx + c, a ¹ 0 is a quadratic polynomial and a, b are two zeroes of polynomial then
v EXAMPLES v
Ex.9 Find the zeroes of the quadratic polynomial
6×2 – 13x + 6 and verify the relation between the zeroes and its coefficients.
Sol. We have, 6×2 – 13x + 6 = 6×2 – 4x – 9x + 6
= 2x (3x – 2) –3 (3x – 2)
= (3x – 2) (2x – 3)
So, the value of 6×2 – 13x + 6 is 0, when
(3x – 2) = 0 or (2x – 3) = 0 i.e.,
When x = or
Therefore, the zeroes of 6×2 – 13x + 6 are
and .
Sum of the zeroes
= + = = =
Product of the zeroes
= × = =
Ex.10 Find the zeroes of the quadratic polynomial
4×2 – 9 and verify the relation between the zeroes and its coefficients.
Sol. We have,
4×2 – 9 = (2x)2 – 32 = (2x – 3) (2x + 3)
So, the value of 4×2 – 9 is 0, when
2x – 3 = 0 or 2x + 3 = 0
i.e., when x = or x = –.
Therefore, the zeroes of 4×2 – 9 are & –.
Sum of the zeroes
= – = 0 = =
Product of the zeroes
= = =
Ex.11 Find the zeroes of the quadratic polynomial
9×2 – 5 and verify the relation between the zeroes and its coefficients.
Sol. We have,
9×2 – 5 = (3x)2 – ()2 = (3x – ) (3x +)
So, the value of 9×2 – 5 is 0,
when 3x – = 0 or 3x + = 0
i.e., when x = or x = .
Sum of the zeroes
= – = 0 = =
Product of the zeroes
= = =
Ex.12 If a and b are the zeroes of ax2 + bx + c, a ¹ 0 then verify the relation between the zeroes and its coefficients.
Sol. Since a and b are the zeroes of polynomial
ax2 + bx + c.
Therefore, (x – a), (x – b) are the factors of the polynomial ax2 + bx + c.
Þ ax2 + bx + c = k (x – a) (x – b)
Þ ax2 + bx + c = k {x2 – (a + b) x + ab}
Þ ax2 + bx + c = kx2 – k (a + b) x + kab …(1)
Comparing the coefficients of x2, x and constant terms of (1) on both sides, we get
a = k, b = – k (a + b) and c = kab
Þ a + b = – and ab =
a + b = and ab = [Q k = a]
Sum of the zeroes = =
Product of the zeroes = =
Ex. 13 Prove relation between the zeroes and the coefficient of the quadratic polynomial
ax2 + bx + c.
Sol. Let a and b be the zeroes of the polynomial
ax2 + bx + c
\ a = ….(1)
b = ….(2)
By adding (1) and (2), we get
a + b = +
= = – =
Hence, sum of the zeroes of the polynomial
ax2 + bx + c is –
By multiplying (1) and (2), we get
ab =
= =
= =
=
Hence, product of zeroes =
Ex.14 find the zeroes of the quadratic polynomial
x2 – 2x – 8 and verify a relationship between zeroes and its coefficients.
Sol. x2 – 2x – 8 = x2 – 4x + 2x – 8
= x (x – 4) + 2 (x – 4) = (x – 4) (x + 2)
So, the value of x2 – 2x – 8 is zero when
x – 4 = 0 or x + 2 = 0 i.e., when x = 4 or x = – 2.
So, the zeroes of x2 – 2x – 8 are 4, – 2.
Sum of the zeroes
= 4 – 2 = 2 = =
Product of the zeroes
= 4 (–2) = –8 = =
Ex.15 Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients. 2×3 + x2 – 5x + 2 ; , 1, – 2
Sol. Here, the polynomial p(x) is
2×3 + x2 – 5x + 2
Value of the polynomial 2×3 + x2 – 5x + 2
when x = 1/2
= 2+– 5+2 =+–+ 2 = 0
So, 1/2 is a zero of p(x).
On putting x = 1 in the cubic polynomial
2×3 + x2 – 5x + 2
= 2(1)3 + (1)2 – 5(1) + 2 = 2 + 1 – 5 + 2 = 0
On putting x = – 2 in the cubic polynomial
2×3 + x2 – 5x + 2
= 2(–2)3 + (–2)2 – 5 (–2) + 2
= – 16 + 4 + 10 + 2 = 0
Hence, , 1, – 2 are the zeroes of the given polynomial.
Sum of the zeroes of p(x)
= + 1 – 2 = – =
Sum of the products of two zeroes taken at a time
= × 1 + × (–2) + 1 × (–2)
= – 1 – 2 = – =
Product of all the three zeroes
= × (1) × (–2) = –1
==
² Symmetric function :
An algebraic expression in a and b, which remains unchanged, when a and b are interchanged is known as symmetric function in a and b.
For example, a2 + b2 and a3 + b3 etc. are symmetric functions. Symmetric function is to be expressed in terms of (a + b) and ab. So, this can be evaluated for a given quadratic equation.
² Some useful relations involving a and b :
- a2 + b2 = (a + b)2 – 2ab
- (a – b)2 = (a + b)2 – 4ab
- a2 – b2 = (a + b) (a – b) = (a + b)
- a3 + b3 = (a + b)3 – 3ab (a + b)
- a3 – b3 = (a – b)3 + 3ab (a – b)
- a^{4} + b^{4} =[(a + b)^{2} – 2ab]^{2} –2(ab)^{2}
- a^{4} – b^{4} = (a^{2} + b^{2}) (a^{2} – b^{2}) then use (1) and (3)
v EXAMPLES v
Ex.16 If a and b are the zeroes of the polynomial
ax2 + bx + c. Find the value of
(i) a – b (ii) a2 + b2.
Sol. Since a and b are the zeroes of the polynomial ax2 + bx + c.
\ a + b = – ; ab =
(i) (a – b)2 = (a + b)2 – 4ab
= – = – =
a – b =
(ii) a2 + b2 = a2 + b2 + 2ab – 2ab
= (a + b)2 – 2ab
= – 2 =
Ex.17 If a and b are the zeroes of the quadratic polynomial ax2 + bx + c. Find the value of
(i) a2 – b2 (ii) a3 + b3.
Sol. Since a and b are the zeroes of ax2 + bx + c
\ a + b = , ab =
(i) a2 – b2 = (a + b) (a – b)
= –
= – = –
= –
(ii) a3 + b3 = (a + b) (a2 + b2 – ab)
= (a + b) [(a2 + b2 + 2ab) – 3ab]
= (a + b) [(a + b)2 – 3ab]
=
==
=
v EXAMPLES v
Ex.18 Form the quadratic polynomial whose zeroes are 4 and 6.
Sol. Sum of the zeroes = 4 + 6 = 10
Product of the zeroes = 4 × 6 = 24
Hence the polynomial formed
= x2 – (sum of zeroes) x + Product of zeroes
= x2 – 10x + 24
Ex.19 Form the quadratic polynomial whose zeroes are –3, 5.
Sol. Here, zeroes are – 3 and 5.
Sum of the zeroes = – 3 + 5 = 2
Product of the zeroes = (–3) × 5 = – 15
Hence the polynomial formed
= x2 – (sum of zeroes) x + Product of zeroes
= x2 – 2x – 15
Ex.20 Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively-
(i) , – 1 (ii) , (iii) 0,
Sol. Let the polynomial be ax2 + bx + c and its zeroes be a and b.
(i) Here, a + b = and a . b = – 1
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – x – 1 = x2 – – 1
The other polynomial are k
If k = 4, then the polynomial is 4×2 – x – 4.
(ii) Here, a + b =, ab =
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – () x + or x2 –x +
Other polynomial are k
If k = 3, then the polynomial is
3×2 – 3x + 1
(iii) Here, a + b = 0 and a.b =
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – (0) x + = x2 +
Ex.21 Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2,
– 7 and –14, respectively.
Sol. Let the cubic polynomial be
ax3 + bx2 + cx + d
Þ x3 + x2 + x + ….(1)
and its zeroes are a, b and g, then
a + b + g = 2 = –
ab + bg + ag = – 7 =
abg = – 14 = –
Putting the values of , and in (1),
we get
x3 + (–2) x2 + (–7)x + 14
Þ x3 – 2×2 – 7x + 14
Ex.22 Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively.
Sol. Let the cubic polynomial be
ax3 + bx2 + cx + d
Þ x3 + x2 + x + ….(1)
and its zeroes are a, b, g. Then
a + b + g = 0 = –
ab + bg + ag = – 7 =
abg = – 6 =
Putting the values of , and in (1),
we get
x3 – (0) x2 + (–7) x + (–6)
or x3 – 7x + 6
Ex.23 If a and b are the zeroes of the polynomials
ax2 + bx + c then form the polynomial whose zeroes are and .
Sol. Since a and b are the zeroes of ax2 + bx + c
So a + b = , ab =
Sum of the zeroes = + =
= =
Product of the zeroes
= ==
But required polynomial is
x2 – (sum of zeroes) x + Product of zeroes
Þ x2 – x +
or x2 + x +
or c
Þ cx2 + bx + a
Ex.24 If a and b are the zeroes of the polynomial
x2 + 4x + 3, form the polynomial whose zeroes are 1 + and 1 + .
Sol. Since a and b are the zeroes of the polynomial x2 + 4x + 3.
Then, a + b = – 4, ab = 3
Sum of the zeroes
= 1 + + 1 + =
= = = =
Product of the zeroes
= = 1 + + +
= 2 + =
= = =
But required polynomial is
x2 – (sum of zeroes) x + product of zeroes
or x2 – x + or k
or 3 (if k = 3)
Þ 3×2 – 16x + 16
Ø
Step 1:
First arrange the term of dividend and the divisor in the decreasing order of their degrees.
Step 2 :
To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor.
Step 3 :
To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.
Step 4 :
Continue this process till the degree of remainder is less than the degree of divisor.
² Division Algorithm for Polynomial
If p(x) and g(x) are any two polynomials with
g(x) ¹ 0, then we can find polynomials q(x) and r(x) such that
p(x) = q(x) × g(x) + r(x)
where r(x) = 0 or degree of r(x) < degree of g(x).
The result is called Division Algorithm for polynomials.
v EXAMPLES v
Ex.25 Divide 3×3 + 16×2 + 21x + 20 by x + 4.
Sol.
Quotient = 3×2 + 4x + 5
Remainder = 0
Ex.26 Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below :
p(x) = x3 – 3×2 + 5x – 3, g(x) = x2 – 2
Sol. We have,
p(x) = x3 – 3×2 + 5x – 3 and g(x) = x2 – 2
We stop here since
degree of (7x – 9) < degree of (x2 – 2)
So, quotient = x – 3, remainder = 7x – 9
Therefore,
Quotient × Divisor + Remainder
= (x – 3) (x2 – 2) + 7x – 9
= x3 – 2x – 3×2 + 6 + 7x – 9
= x3 – 3×2 + 5x – 3 = Dividend
Therefore, the division algorithm is verified.
Ex.27 Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below
p(x) = x4 – 3×2 + 4x + 5, g (x) = x2 + 1 – x
Sol. We have,
p(x) = x4 – 3×2 + 4x + 5, g(x) = x2 + 1 – x
We stop here since
degree of (8) < degree of (x2 – x + 1).
So, quotient = x2 + x – 3, remainder = 8
Therefore,
Quotient × Divisor + Remainder
= (x2 + x – 3) (x2 – x + 1) + 8
= x4 – x3 + x2 + x3 – x2 + x – 3×2 + 3x – 3 + 8
= x4 – 3×2 + 4x + 5 = Dividend
Therefore the Division Algorithm is verified.
Ex.28 Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm.t2 – 3; 2t4 + 3t3 – 2t2 – 9t – 12.
Sol. We divide 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3
Here, remainder is 0, so t2 – 3 is a factor of
2t4 + 3t3 – 2t2 – 9t – 12.
2t4 + 3t3 – 2t2 – 9t – 12 = (2t2 + 3t + 4) (t2 – 3)
Ex.29 Obtain all the zeroes of
3×4 + 6×3 – 2×2 – 10x – 5, if two of its zeroes are and – .
Sol. Since two zeroes are and –,
x = , x = –
Þ = x2 – or 3×2 – 5 is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial and 3×2 – 5.
So, 3×4 + 6×3 – 2×2 – 10x – 5
= (3×2 – 5) (x2 + 2x + 1) + 0
Quotient = x2 + 2x + 1 = (x + 1)2
Zeroes of (x + 1)2 are –1, –1.
Hence, all its zeroes are , – , –1, –1.
Ex.30 On dividing x3 – 3×2 + x + 2 by a polynomial
g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).
Sol. p(x) = x3 – 3×2 + x + 2
q(x) = x – 2 and r (x) = –2x + 4
By Division Algorithm, we know that
p(x) = q(x) × g(x) + r(x)
Therefore,
x3 – 3×2 + x + 2 = (x – 2) × g(x) + (–2x + 4)
Þ x3 – 3×2 + x + 2 + 2x – 4 = (x – 2) × g(x)
Þ g(x) =
On dividing x3 – 3×2 + 3x – 2 by x – 2,
we get g(x)
Hence, g(x) = x2 – x + 1.
Ex.31 Give examples of polynomials p(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg q(x) = 0
Sol. (i) Let q(x) = 3×2 + 2x + 6, degree of q(x) = 2
p(x) = 12×2 + 8x + 24, degree of p(x) = 2
Here, deg p(x) = deg q(x)
(ii) p(x) = x5 + 2×4 + 3×3 + 5×2 + 2
q(x) = x2 + x + 1, degree of q(x) = 2
g(x) = x3 + x2 + x + 1
r(x) = 2×2 – 2x + 1, degree of r(x) = 2
Here, deg q(x) = deg r(x)
(iii) Let p(x) = 2×4 + 8×3 + 6×2 + 4x + 12
q(x) = 2, degree of q(x) = 0
g(x) = x4 + 4×3 + 3×2 + 2x + 6
r(x) = 0
Here, deg q(x) = 0
Ex.32 If the zeroes of polynomial x^{3} – 3x^{2} + x + 1 are a – b, a , a + b. Find a and b.^{ }
Sol. Q a – b, a, a + b are zeros
\ product (a – b) a(a + b) = –1
Þ (a^{2} – b^{2}) a = –1 …(1)
and sum of zeroes is (a – b) + a + (a + b) = 3
Þ 3a = 3 Þ a = 1 …(2)
by (1) and (2)
(1 – b^{2})1 = –1
Þ 2 = b^{2} Þ b = ±
\ a = –1 & b = ± Ans.
Ex.33 If two zeroes of the polynomial
x^{4} – 6x^{3} –26x^{2} + 138x – 35 are 2 ± ,
find other zeroes.
Sol. Q 2 ±are zeroes.
\ x = 2 ±
Þ x – 2 = ±(squaring both sides)
Þ (x – 2)^{2} = 3 Þ x^{2} + 4 – 4x – 3 = 0
Þ x^{2} – 4x + 1 = 0 , is a factor of given polynomial
\ other factors
=
\ other factors = x^{2} – 2x – 35
= x^{2} – 7x + 5x – 35 = x(x – 7) + 5(x – 7)
= (x – 7) (x + 5)
\ other zeroes are (x – 7) = 0 Þ x = 7
x + 5 = 0 Þ x = – 5 Ans.
Ex.34 If the polynomial x^{4} – 6x^{3} + 16x^{2} –25x + 10 is divided by another polynomial x^{2} –2x + k, the remainder comes out to be x + a, find k & a.
Sol.
According to questions, remainder is x + a
\ coefficient of x = 1
Þ 2k – 9 = 1
Þ k = (10/2) = 5
Also constant term = a
Þ k^{2} – 8k + 10 = a Þ (5)^{2} – 8(5) + 10 = a
Þ a = 25 – 40 + 10
Þ a = – 5
\ k = 5, a = –5 Ans.
exercise # 1
Factorize each of the following expression
Q.1 x^{2 }– x – 42
Q.2 6 – 5y – y^{2}
Q.3 a^{2 }+ 46a + 205
Q.4 ab + ac – b^{2} – bc
Q.5 p^{4} – 81q^{4}
Use remainder theorem to find remainder,
when p(x) is divided by q(x) in following questions.
Q.6 p(x) = 2x^{2 }– 5x + 7, q(x) = x – 1
Q.7 p(x) = x^{9} – 5x^{4} + 1 , q(x) = x + 1
Q.8 p(x) = 2x^{3} – 3x^{2} + 4x – 1 , q(x) = x + 2
Q.9 Find positive square root of 36x^{2} + 60x + 25
Q.10 Simplify :
Q.11 (x^{2} + 4y)^{2} + 21 (x^{2 }+ 4y) + 98
Q.12 Find the value of k if (x – 2) is a factor of
2x^{3}– 6x^{2}+ 5x + k.
Q.13 Find the value of k if (x + 3) is a factor of
3x^{2} + kx + 6.
Q.14 p(x) = 3x^{6 }– 7x^{5} + 7x^{4 }– 3x^{3} + 2x^{2} – 2, q(x) = x – 1
Q.15 For what value of k is y^{3} + ky + 2k – 2 exactly divisible by (y + 1) ?
Q.16 If x + 1 and x – 1 are factors of
mx^{3} + x^{2} – 2x + n, find the value of m and n.
Q.17 Find the zeros of the polynomial
f(x) = 2x^{2} + 5x – 12 and verify the relation between its zeroes and coefficients.
Q.18 Find the zeroes of the polynomial f(x) = x^{2} – 2 and verify the relation between its zeroes and coefficients.
Q.19 Obtain the zeroes of the quadratic polynomial x^{2} – 8x + 4and verify the relation between its zeroes and coefficients.
Q.20 Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as
2, –7 and –14 respectively.
Q.21 Find a cubic polynomial whose zeroes are 3, 5 and – 2.
Q.22 Divide 5x^{3} – 13x^{2} + 21x – 14 by (3 – 2x + x^{2}) and verify the division algorithm.
Q.23 What real number should be subtracted from the polynomial (3x^{3} + 10x^{2 }– 14x + 9) so that (3x – 2) divides it exactly?
Q.24 Find all the zeroes of (2x^{4} – 3x^{3} – 5x^{2} + 9x –3), it being given that two of its zeroes are
and – .
answer key
- Verty Short Answer Type :
- (x + 6) (x – 7) 2. (6 + y) (1 – y) 3. (a + 41) (a + 5) 4. (a – b) (b + c)
- (p + 3q) (p – 3q) (p^{2 }+ 9q^{2}) 6. 4 7. – 5 8. –37
- Short Answer Type :
- 6x + 5 10. (a +b) 11. (x^{2 }+ 4y + 7) (x^{2 }+ 4y + 14) 12. –2
- 11 15. 3
- Long Answer Type :
- m = 2, n = – 1 17. –4, 18. 19.
- x^{3} – 2x^{2} – 7x + 14 21. x^{3} – 6x^{2} – x + 30 22. quotient = 5x – 3, Remainder = –5
- 5 24.
exercise # 2
Q.1 If = 3, then find value of .
Q.2 If =, then find .
Q.3 If = 4, then find .
Q.4 If (x – 2) is a factor of (x^{2} + 3qx – 2q), then find the value of q.
Q.5 If x^{3} + 6x^{2} + 4x + k is exactly divisible by
(x + 2), then find the value of k.
Q.6 Let f(x) = x^{3} – 6x^{2} + 11x – 6. Then, which one of the following is not factor of f(x) ?
(A) x – 1 (B) x – 2
(C) x + 3 (D) x – 3
Q.7 If x^{100} + 2x^{99} + k is divisible by (x + 1), then find the value of k.
Q.8 On dividing (x^{3} – 6x + 7) by (x + 1), find the remainder.
Q.9 Find the value of expression (16x^{2 }+ 24x + 9) for x = – .
Q.10 If 2x^{3} + 5x^{2} – 4x – 6 is divided by 2x + 1, then find remainder.
Q.11 If p(x) = x^{2} – 2x – 3, then find
(i) p(3); (ii) p(–1)
Q.12 Find the zeros of the quadratic polynomial
(6x^{2} – 7x – 3) and verify the relation between its zeros and coefficients.
Q.13 Find the zeros of the quadratic polynomial
(5u^{2} + 10u) and verify the relation between the zeros and the coefficients.
Q.14 Find the quadratic polynomial whose zeros are and . Verify the relation between the coefficients and the zeros of the polynomial.
Q.15 Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.
Q.16 Find the quadratic polynomial , the sum of whose zeros is –5 and their product is 6. Hence, find the zeros of the polynomial.
Q.17 Find the quadratic polynomial, the sum of whose zeros is 0 and their product is –1. Hence, find the zeros of the polynomial.
Q.18 Find a quadratic polynomial whose one zero is 5 + .
Q.19 On dividing (x^{3} – 3x^{2} + x + 2) by a polynomial g(x), the quotient and remainder are (x – 2) and (–2x + 4) respectively. Find g(x).
Q.20 If the polynomial (x^{4} + 2x^{3} + 8x^{2} + 12x + 18) is divided by another polynomial (x^{2} + 5), the remainder comes out to be (px + q). Find the value of p and q.
Q.21 Obtain all zeros of the polynomial
(2x^{3} – 4x – x^{2} + 2), if two of its zeros are and –.
Q.22 If 1 and –2 are two zeros of the polynomial
(x^{3} – 4x^{2} – 7x + 10), find its third zero.
Q.23 Find all the zeros of the polynomial
(2x^{4} – 11x^{3} + 7x^{2} + 13x – 7), it being given that two if its zeros are (3 + ) and (3 – ).
Q.24 If a, b are the zeros of the polynomial
f(x) = x^{2} – 5x + k such that a – b = 1, find the value of k.
Q.25 Show that the polynomial f(x) = x^{4} + 4x^{2} + 6 has no zero.
Q.26 Use remainder theorem to find the value of k, it being given that when x^{3} + 2x^{2} + kx + 3 is divided by (x – 3), then the remainder is 21.
answer key
- 7 2. 9 4. – 1 5. – 8 6. (C)
- 1 8. 12 9. 0 10. – 3
- (i) 0 , (ii) 0 12. 13. – 2, 0 14. 12x^{2} – 5x – 2
- (x^{2} – 8x + 12), {6, 2} 16. (x^{2} + 5x + 6), {– 3, – 2}
- (x^{2} – 1), {1, –1} 18. x^{2} – 10x + 18 19. x^{2} – x + 1 20. p = 2, q = 3
- 22. 5 23. , , – 1
- k = 6 26. k = – 9