Factor and Remainder Theorems

Edexcel IAL WMA12/01/P2/Oct 2021/Q04 (Remainder and Factor Theorem)

f(x) = (x2 - 2)(2x - 3) - 21 

• State the value of the remainder when f(x) is divided by (2x − 3)​​ 

(1)​​ 

• Use the factor theorem to show that (x − 3) is a factor of f(x)​​ 

(2)​​ 

• Hence,​​ 

• • factorise f(x)​​ 

  • show that the equation f(x) = 0 has only one real root.​​ 

(5)

SOLUTION

a-​​ 

METHOD # 1 (Recommended)

Remember, there is a short and easier method to know the value of the remainder directly from the equation of the function. The constant term outside the product is directly given by the constant term outside the product, which in this question is ​​ −21.

So, the remainder is −21.

METHOD # 2 (Not worthy as the question is of just 1 mark)

fx=(x2 - 2)(2x - 3) - 21 

f32=322 - 2×232- 3   - 21 

f32=-21

Hence, the remainder= -21 

b-

f(x) = (x2 - 2)(2x - 3) - 21

Using the factor theorem.​​ 

If x= 3 is a root then f3 must be equals to​​ 0.

f3=(32 - 2)(2×3 - 3) – 21

f3=7 x 3-21

f3=0

Hence,​​ x-3​​ is a factor of​​ f(x).

c- i- Lets factorise f(x) using long division method.

fx=x2-22x-3-21

=2x3-3x2-4x+6-21

fx=2x2-3x2-4x-15

Since​​ x-3 is a factor, so lets divide f(x).

2x2+3x+52x3-3x2-4x-152x3-6x2             x-3         3x2-4x        -3x2-9x                                  5x-15                                  5x-150         0 

fx=x-32x2+3x+5

c-ii- ​​ To show that the equation f(x) = 0 has only one real root, lets solve for the equation

fx=0

 x-32x2+3x+5=0

x-3=0           2x2+3x+5=0 

x=3           2x2+3x+5=0

To solve​​ 2x2+3x+5=0, lets find the discriminant,​​ b2-4ac, where​​ a=2,​​ b=3, and​​ c=5.

b2-4ac=32-425

=9-40

=-31<0   (no real root)

As ​​ b2-4ac<0, therefore are no real roots to​​ 2x2+3x+5.​​ 
Hence, ​​ x=3​​ is the only red root of f(x).