Factor and Remainder Theorems

Edexcel IAL WMA12/01/P2/June 2022/Q07 (The Factor Theorem, Integration, Simultaneous Equation)

f(x) = Ax3 + 6x2 - 4x + B 

where A and B are constants.​​ 

Given that​​ 

• (x + 2) is a factor of f(x)​​ 

• ∫35fxdx=176 

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

(7)

SOLUTION​​ 

This question is a combined topic question, from integration, and algeabraic expression.

First, we are going to divide the cubic equation by the factor​​ (x+2), and equate it to 0 since it must satisfy the equation. This would give us the equation in terms of A and B. Then we will solve the integration and make the second equation. Hence, we will solve both equations simultaneously to get the value of A and B.​​ 

If​​ x+2 ​​​​ is a factor ​​ of​​ f2, then​​ f-2=0.​​ 

So,​​ 

f(x) = Ax3 + 6x2 - 4x + B

0=A-23+6-22-4-2+B

-8A+24+8+B=0

-8A+B+32=0

-8A+B= -32         -1

Now, solving the​​ ∫35fxdx, and equation it to 176.

∫35fx dx=176

∫35Ax3+x2-4x+Bdx=176

Ax44+6x33-4x22+Bx35=176

Ax44+2x3+2x2+Bx35 =176

Applying the limits.​​ 

A544+253-252+B5-A344+233-232+B3 =176

625A4+250-50+5B-81A4+54-18+3B=176

136A+200+5B-36+3B=176

136A+164+2B=176

136A+2B= 12

68A+B=6            -2

Now, solving equation 1 and 2 simultaneously. On subtracting equation 2 from 1, B gets eliminated.​​ 

-8A+B= -32-68A+B=6-72A = -38

A= -30-76

A=12 

​​ And, the value of B is found by substituting the value of A in either equation 1 or 2.​​ 

Here, we are using equation 2:

68A+B=6            -2

6812+B=6 

B=6-34

B=-28