Nine Standard >> Factor theorem and factorization using Factor theorem

Factor theorem:

Let P(x) be a polynomial and 'a' be a real number then x-a is a factor of P(x) if P(a)=0, i.e 
(1) If (x-a) is a factor of P(x), then P(a)=0
(2) If P(a)=0, then (x-a) is a factor of P(x).

Example: \(Let P(x)=x^3+3x^2+3x+1\) check whether x+1 and \(x-\frac{1}{2}\) are factors of P(x).

Solve:
divisor=x+1, a=-1
\(P(-1)=(-1)^3+3(-1)^2+3(-1)+1\)
        =-1+3-3+1=0

So x+1 is a factor of P(x)
    \(P(\frac{1}{2})=(\frac{1}{2})^3+3(\frac{1}{2})^2+3(\frac{1}{2})+1 \neq 0\)
\(x-\frac{1}{2}\) is not a factor P(x)

Problem:

Find the value of k if x-1 is a factor of \(4x^3+3x^2-4x+k\)

Solve:

\(P(x)=4x^3+3x^2-4x+k\)
        
Divisor=x-1 and it factor 
So, according to the factor theorem remainder 

P(1)=0  ---(1)

The value of P(1)=
\(4.(1)^3+3(1)^2-4(1)+k\)---(2)
Form equation (1) and (2) we get
\(4.(1)^3+3(1)^2-4(1)+k=0\)
or, 4+3-4+k=0
or, 3-k=0
So, k=-3