Factor and Remainder Theorems

Edexcel IAL WMA12/01/P2/Jan 2023/Q05 (Remainder & Factor Theorem, Algebraic Division)

f(x) = x3 +  p + 3x2 – x + q

where p and q are constants and​​ p>0​​ 

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

• show that​​ 

9p + q = –51

(2)

Given also that when f (x) is divided by (x + p) the remainder is 9​​ 

• show that​​ 

3p2 + p + q – 9 = 0

(2)

• Hence find the value of p and the value of q.​​ 

(3)

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

fx= x – 3gx

(2)

SOLUTION​​ 

a- ​​ Using the factor theorem, show that​​ x-3is a factor of fx.

(Here, in our case, we will use the second point of the factor theorem that is since​​ x-3​​ is a factor of

​​ fx​​ then​​ f3=0)​​ 

f(x) = x3 +  p + 3x2 – x + q

∴f3=0

33+p+332-3+q=0

27+9p+3-3+q=0

279p+27-3+q=0

9p+q+51=0 

9p+q= -51

b- ​​ Using the factor theorem, to show the given equation in the question.​​ 

(Again, we will use factor theorem, but this time the remainder is said to be 9 when​​ f(x)​​ is divided by​​ (x+p), which means that​​ f-p=9.)

f(x) = x3 +  p + 3x2 – x + q

∴f-p=9

-p3+p+3-p2--p+q=9

-p3+p2p+3+p+q=9 

-p3+p3+3p2+p+q=9

3p2+p+q-9=0 

c- ​​ Solving equations from part (a) and (b) simultaneously to get the value of p and q.​​ 

(Using substitution method to solve simultaneous equation that is making q as a subject from equation of we get in part (a) and substituting the value of q in equation we get from part (b).)

From part (a), we have

9p+q= -51

Making​​ q​​ as the subject of the equation.​​ 

q= -51-9p

Now, substituting in equation​​ 

3p2+p+q-9=0 

3p2+p+-51-9p-9= 0

3p2+p-51-9p-9=0

3p2-8p-60=0

3p+10p-6=0

3p+10=0      p-6=0

p=-103   &  p=6

Since​​ P>0, reject negative value of p.

Substituting the value of​​ p​​ in

q= -51-9p

q= -51-  96= -51-54= -105

q=-105  

Hence, the value of​​ p​​ and​​ q​​ is​​ 6​​ and​​ -105​​ respectively.

d- Using long division method.

(We can use long division method to divide a polynomial by (x±p), where p is the constant.)

fx=(x-3)g(x)

Where,​​ 

fx=x3+9x2-x-105 

x3+9x2-x-105=x-3gx

g(x)=x3+9x2-x-105x-3

                 x2+12x+35(x-3) x3+9x2-x-105-x3-3x2             12x2-x            -12x2-36x                 35x-105                 35x-105                 0

So after long division we get.​​ 

(Remember,​​ fxx±p=quotient, and in our case, quotient is x2+12x+35).

x3+9x2-x-105x-3=x2+12x+35

Hence,​​ 

gx=x2+12x+35