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Factor Theorem

Types of Polynomials Assignment / Homework Help
Let p(x) be a polynomial of degree n ≥ 1 and a be any real number. Then by factor theorem, x a is a factor of p(x), if p(a) = 0 and p(a) = 0, if x a is a factor of p(x).

FACTORISATION OF THE QUADRATIC POLYNOMIAL OF TYPE x2 + lx + m:

For a quadratic polynomial of type x2 + lx + m, where l, m are constants, first we split the middle term lx as ax + bx so that ab = m, then we get

x2 + lx + m = x2 + ax + bx + ab
= x(x + a) + b(x + a)
= (x + a)(x + b)

FACTORISATION OF THE QUADRATIC POLYNOMIAL OF TYPE ax2 + bx + c:

For a quadratic polynomial of type axx2 + bx + c, where a ≠ 0 and a, b, care constants, first we assume that (px + q) and (rx + s) are its factors. Then we get,

ax2 + bx + c = (px + q) (rx + s)
= prx2 + (ps + qr)x + qs

Now comparing the coefficients of x2 on both sides we get,

a = pr

and comparing the coefficients of x on both sides we get,

b = ps + qr

Comparing the constant terms we get,

c = qs

Hence this implies that b is a sum of two numbers ps and qr, and the product of these two numbers (ps)(qr) = (pr)(qs) = ac.

Hence in order to factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ac.

Example 1:

Using the factor theorem determine whether x + 1 is a factor of the polynomial 2x3 + x2 2x - 1 or not.

Solution:

Let p(x) = 2x3 + x2 2x 1

Now x + 1 = 0 x = -1

Hence p(-1) = 2(-1)3 + (-1)2 2(-1) 1
= -2 + 1 + 2 1
= 0

Hence by factor theorem, x + 1 is a factor of p(x).


Example 2:

Factorise x2 5x + 6.

Solution:

Splitting the middle term we get:

x2 - 5x + 6 = x2 3x 2x + 6
= x(x 3) 2(x 3)
= (x- 3)(x 2)            (Answer)

Hence by factor theorem, x + 1 is a factor of p(x).


Example 3:

Factorise 12x2 7x + 1.

Solution:

First we find the product of 12 so that the sum of the two numbers is 7. i.e., 12 = 43 (also 4 + 3 = 7)

Hence 12x2 7x + 1 = 12x2 4x 3x + 1
= 4x(3x 1) -1(3x 1)
= (3x 1)(4x 1)        (Answer)


Example 4:

Find the values of p and q so that x + 2 and x - 1 are factors of the polynomial x3 + 10x2 + px + q.

Solution:

Let g(x) = x3 + 10x2 + px + q

Now given that x + 2 is a factor of g(x).

By factor theorem if x + 2 is a factor of g(x), then g(-2) = 0.

Hence g(-2) = (-2)3 + 10(-2)2 + p(-2) + q = 0
= -8 + 40 -2p + q = 0
= 2p q = 32                          (1)

Again if x 1 is a factor of g(x), then g(-2) = 0.
g(1) = 0

Hence (1)3 + 10(1)2 + p(1) + q = 0
=> 1 + 10 + p + q = 0
=> p + q = -11                          (2)

Adding (1) and (2) we get,

3p = 21
=> p = 7

Putting the value of p = 7 in (2) we get,
7 + q = -11
=> q = -18

Hence p = 7 and q = -18               (Answer)


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