Factor Theorem

Statement

If p(x), a polynomial in x is divided by x-a and the remainder = f (a) is zero, then (x-a) is a factor of p(x).

Proof:

When p(x) is divided by x-a,

R = p(a) (by remainder theorem)

p(x) = (x-a).q(x)+p(a)

(Dividend = Divisor x quotient + Remainder Division Algorithm)

But p(a) = 0 is given.

Hence p(x) = (x-a).q(x)

Conversely if x-a is a factor of p(x) then p(a)=0.

p(x) = (x-a).q(x) + R

If (x-a) is a factor then the remainder should be zero (x-a divides p(x)

exactly)

R=0

By remainder theorem, R = p(a)

p(a)=0

Illustrative Examples

Example 1:

Determine whether x-2 is a factor of x²-7x+10.

Suggested answer:

p(x) = x²-7x+10 is divided by x-2.

R = p(2)

= 4-14+10=0, R=0

Example 2:

Determine whether x-3 is a factor of x³-3x²+4x-12.

Suggested answer:

p(x) = x³-3x²+4x-12 is divided by x-3.

R = p(3)

= 3³- 3 x 3² + 4 x 3-12

= 27-27+12-12=0

Example 3:

Show that x+1 is a factor of 2x³+5x²-9x-12.

Suggested answer:

Let p(x) = 2x³+5x²-9x-12

when p(x) is by x+1 [x- (-1)],

R = p(-1)

= 2(-1)³+5(-1)²-9(-1)-12

= -2+5+9-12

= -14+14 = 0

R = 0

Example 4:

Find a so that x⁴+2x³-ax²+x-2 has (x+2) as its factor.

Suggested answer:

Let p(x) = x⁴+2x³-ax²+x-2

when p(x) is by x+2 [x- (-2)],

R = p(-2)

= (-2)⁴+2(-2)³-a(-2)²-2-2

= 16-16-4a-4

= -4a - 4

As (x+2) is a factor, R=0

-4a - 4 = 0

a = -1

Example 5:

If (x-2) and (x-3) are factors of x³+ax²+bx+12, find a and b.

Suggested answer:

Let p(x) = x³+ax²+bx+12

when p(x) is x-2,

R = p(2)

= 2³+a(2)²+b(2)+12

= 8+4a+2b+12

= 20+4a+2b

But x-2 is a factor,

20+4a+2b = 0

when p(x) is by x-3,

R = p(3)

= 27+9a+3b+12

= 39+9a+3b

But x-3 is a factor,

39+9a+3b=0

[9a+3b=-39] 3

3a+b = -13 …(2)

Subtracting equation (1) from (2), we get,

3a+b-2a-b = -13+10

a = -3

Substitute a = -3 in equation (1),

2a+b = -10

-6+b = -10

b = -4

Example 6:

Factorise using factor theorem.

a) 2x²+x-3

b) 5x²+6x-8

Suggested answer:

a) 2x²+x-3

Let p(x) = 2x²+x-3

The factors of the constant 3 are +1, -1, +3, -3.

Let us find p(1) = 2+1-3 = 0

Divide 2x²+x-3 by x-1 to get the other factor.

Hence 2x²+x-3 = (2x+3) (x-1) .

b) 5x²+6x-8

Suggested answer:

Let p(x) = 5x²+6x-8

Factors of the constant '8' are 1,-1,2,-2,4, -4 and 8,-8.

Let us find the value of p(x) at each one of them to get one factor.

p(-2) = 20-12-8 = 0

Divide 5x²+6x-8 by x+2 to get the other factor.