Factor and Remainder Theorems

WMA12/01/IAL/Edexcel/P2/Oct 2019/Q4 (Factor & Remainder Theorems)

f(x) = (x - 3)(3x2 + x + a) - 35​​ where a is a constant

• • State the remainder when f(x) is divided by (x - 3)

(1)

Given (3x - 2) is a factor of f(x)

• • show that a = - 17

(2)

(c) ​​ Using algebra and showing each step of your working, fully factorise f(x)

(5)

SOLUTION​​ 

a-​​ As​​ (3x-2)​​ is a factor of​​ f(x), the remainder is zero.​​ 

b-​​ Using factor theorem, to find the value of​​ a.

(Consider the second point of the flash card above, so on comparing​​ 3x-2​​ to​​ ax-b,​​ a=3​​ and​​ b=2.)​​ 

3x-2=0

x=23

Since (3x-2) is a factor of ​​ f(x), ​​ f23=0

f(x) = (x - 3)(3x2 + x + a) - 35

f23=0

23-3 3232+23+a-35=0

-73349+23+a=35 

349+23+a=35×-37

43+23+a=-15

2+a= -15

a= -17

Hence, shown.

c- Since​​ f(x) = (x - 3)(3x2 + x + a) – 35, where​​ a=-17.

f(x) = (x - 3)(3x2 + x + a) – 35

f(x) = (x - 3)(3x2 + x-17) – 35

fx=3x3+x2-17x-9x2-3x+51-35

fx=3x3-8x2-20x+16

Now, performing long division method, dividing ​​ 3x3-8x2-20x+16​​ by​​ (x - 3).

x2-2x-8 3x-2⟌3x2-8x2-20x+16

-3x3-2x2                                -6x2-20x                           --6x2+4x                                                        -24x+16                                                         -24x+16 ---     

Thus,​​ 

fx=3x3-8x2-20x+16=3x-2x2-2x-8

Where,​​ x2-2x-8​​ is the quadratic expression, lets factorize it.​​ 

x2-2x-8=x2-4x+2x-8

=xx-4+2x-4

=x+2x-4

Hence,​​ 

fx=3x3-8x2-20x+16=3x-2x2-2x-8

f(x)=3x-2x+2x-4