Factor and Remainder Theorems

Edexcel IAL WMA12/01/P2/Jan 2022/Q05 (Remainder & Factor Theorem, Algebraic Division)

f(x) = 3x3 + Ax2 + Bx – 10 

where A and B are integers.​​ 

Given that​​ 

• when f(x) is divided by (x – 1) the remainder is k​​ 

• when f(x) is divided by (x + 1) the remainder is –10k​​ 

• k is a constant​​ 

• show that​​ 

11A + 9B = 83 

(3)​​ 

Given also that (3x – 2) is a factor of f(x),​​ 

• find the value of A and the value of B.​​ 

(3)​​ 

• Hence find the quadratic expression g(x) such that​​ 

f(x) = 3x – 2gx 

(2)

SOLUTION​​ 

a- Applying the given information that when​​ f(x)​​ is divided by​​ (x – 1)​​ the remainder is​​ k​​ 

f1=313+A12+B1-10

k=313+A12+B1-10

k=3+A+B-10

A+B-7=k

A+B=k+7    -1

Applying the given second ​​ information that when​​ f(x)​​ is divided by​​ (x + 1)​​ the remainder is​​ –10k​​ 

F-1=-13+A-12+B-1-10

-10k=- 3+A-B-10

A-B-13= -10k

A-B=-10k+13     -2

Multiplying the first equation with 10 and solving both equations simultaneously.​​ 

A-B= -10 k+13 10A+10B=10k+7011A+9B=83

Hence, we get​​ 11A+98=83.

b- ​​ The entire part revolves around the application of factor theorem.​​ 

Since​​ (3x-2)​​ is a factor of​​ f(x), it must satisfy the equation of polynomial that is​​ f23=0. (4th​​ point of the above flash card)

f23=0

3233+A232 +B23-10=0

3827+A 49+2B3=10

89+4A9+2B3=10

8+4A+6B=90

4A+6B=82

2A+3B=41

Now, solving the above equation simultaneously with the equation we got in part a. Since both equations are in terms of A and B, by solving simultaneous equations we will be able to get the value of A and B.​​ 

11A+9B=83    1

On multiplying​​ 2A+3B=41​​ with 3 we get​​ 6A+9B=123. So, solving simultaneously.

11A+9B= 83 6A+9B=1235A= -40

5A= -40 

A= -8

Substituting the value of A in anyone of the equation. We have taken​​ 

2A+3B=41

2-8+3B=41

-16+3B=41

3B=57

B=573

B=19

c- ​​ it is given that​​ 

f(x) = (3x – 2) g(x)

3x3-8x2+19x-10=3x-2ax2+bx+c

x2-2x+5

3x3-p8x2+19x-103x-2-3x3-2x2                      -6x2+19x--6x2+4x                    15x-10               15x-10 x         x           x 

Hence, we the quadratic expression is​​ 

gx=x2-2x+5