10 Math Polynomials
10 Math Polynomials
10 Math Polynomials
POLYNOMIALS
Introduction
1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomi-
als respectively.
2. A quadratic polynomial in x with real coefficients is of the form ax2 + bx + c,
where a, b, c are real numbers with a ≠ 0.
3. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points,
where the graph of y = p(x) intersects the x -axis.
4. A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can
have at most 3 zeroes.
5. If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then
b
α + β = −
a
c
αβ =
a
6. If α, β, γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d = 0, then
b
α + β +γ = −
a
c
αβ + βγ + αγ =
a
d
αβγ = −
a
7. The division algorithm states that given any polynomial p(x) and any non-zero
polynomial g(x), there are polynomials q(x) and r(x) such that
EXERCISE 1
1. The graphs of y = p(x) are given below, for some polynomials p(x). Find the num-
ber of zeroes of p(x), in each case.
(i)
Answer: Number of zeroes = 0, since the line is not intersecting the x-axis
(ii)
Answer: Number of zeroes = 1, since the line intersects the x axis once.
(iii)
Answer: Number of zeroes = 2, since the line intersects the x axis twice.
(iv)
(v)
(vi)
EXERCISE 2
1. Find the zeroes of the following quadratic polynomials and verify the rela-
tionship between the zeroes and the coefficients.
(i) x2 – 2x – 8
Answer: x² - 2x – 8
= x² - 4x + 2x – 8
= x (x-4) +2(x-4)
= (x+2)(x-4)
Hence x = -2
Or, x = 4
(ii) 4s2 – 4s + 1
− b± b 2 − 4ac
Answer: Zero =
2a
4± 16 − 4 × 4 × 1
=
8
4 1
= =
8 2
c
Verify: αβ =
a
1 1 1
Or, × = (LHS & RHS are equal)
2 2 4
(iii) 6x2 – 3 – 7x
Answer= 6x² - 7x – 3
− b± b 2 − 4ac
Zero =
2a
7± 49 + 72
=
12
7 ± 121
=
12
7 + 11 18 3
α = = =
12 12 2
7 − 11 4 1
β = = − = −
12 12 3
(iv) 4u2 + 8u
− b± b 2 − 4ac
Answer: Zero =
2a
− 8± 64 + 4 × 4 × 0
=
8
− 8± 8
=
8
− 8+ 8
α = = 0
8
− 8− 8
β = = −2
8
(v) t2 – 15
− b± b 2 − 4ac
Answer: Zero =
2a
15 ± 15
=
2
α = 0, β = 15
(vi) 3x2 – x – 4
4
Zeroes = -1 &
3
2. Find a quadratic polynomial each with the given numbers as the sum and
product of its eroes respectively.
1
(i) ,1
4
Quadratic equation can be written as follows:
(ii) 2,1
Equation: x2 − 2x + 1
(iii) 0, 5
Equation: x 2 − 0x + 5 = x2 + 5
(iv) 1, 1
Equation: x2 − x + 1
1 1
(v) − ,
4 4
1 1
Equation: x2 + x + = 4x 2 + x + 1
4 4
(vi) 4, 1
Equation: x 2 − 4x + 1
EXERCISE 3
1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient
and remainder in each of the following :
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Answer: )
x 2 − 2 x 3 − 3 x 2 + 5 x − 3( x − 3
x3 − 2x
− 3x 2 + 7 x − 3
− 3x 2 + 6
0 + 7x − 9
Quotient: x − 3 , Remainder: 7 x − 9
Answer: ) (
x 2 + 1 − x x 4 − 3x 2 + 4 x + 5 x 2 + x − 3
x4 + x2 − x3
x3 − 4x 2 + 4x + 5
x3 − x 2 + x
− 3 x 2 + 3x + 5
− 3 x 2 + 3x − 3
0+ 0+ 8
Quotient: x 2 + x − 3 , Remainder: 8
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
2 4
)
Answer: − x + 2 x − 5 x + 6 − x − 2
2
(
x 4 − 2x 2
2 x 2 − 5x + 6
2x 2 − 4
− 5 x + 10
Quotient: − x 2 − 2 , Remainder: − 5 x + 10
Answer: ) (
t 2 − 3 2t 4 + 3t 3 − 2t 2 − 9t − 12 2t 2 + 3t + 4
2t 4 − 6t 2
3t 3 + 4t 2 − 9t − 12
3t 3 − 9t
4t 2 − 12
4t 2 − 12
0000000
Since there is no remainder so t 2 − 3 is a factor of the given equation.
Answer: ) (
x 2 + 3 x + 1 3x 4 + 5 x 3 − 7 x 2 + 2 x + 2 3 x 2 − 4 x + 2
3x 4 + 9 x 3 + 3x 2
− 4 x 3 − 10 x 2 + 2 x + 2
− 4 x 3 − 12 x 2 − 4 x
2x 2 + 6x + 2
2x 2 + 6x + 2
0000000000
Since there is no remainder so x 2 + 3 x + 1 is a factor of the given equation
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Answer: ) (
x 3 − 3x + 1 x 5 − 4 x 3 + x 2 + 3x + 1 x 2 − 1
x 5 − 3x 3 + x 2
− x 3 + 3x + 1
− x 3 + 3x − 1
2
3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes
5 5
are and −
3 3
5 5 5
Answer: x+ x − = x 2 − ⇒ 3x 2 − 5 is a factor for the given equation.
3
3 3
) (
3 x 2 − 5 3 x 4 + 6 x 3 − 2 x 2 − 10 x − 5 x 2 + 2 x +1
3x 4 − 5x 2
6 x 3 + 3 x 2 − 10 x − 5
6 x 3 − 10 x
3x 2 − 5
3x 2 − 5
0000000
(
x − 2 ) x 3 − 3 x 2 + 3x − 2 x 2 − x + 1
x 3 − 2x 2
− x 2 + 3x − 2
− x 2 + 2x
x− 2
x− 2
0000
So, g ( x) = x 2 − x + 1