Nine Standard >> Factor theorem and factorization using Factor theorem

Factor theorem:

The Factor Theorem is a fundamental concept in algebra that helps determine whether a given binomial expression is a factor of a polynomial. It provides a method to check whether a specific value is a root (zero) of a polynomial and, consequently, whether the corresponding binomial is a factor of the polynomial.

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)
    \(P(\frac{1}{2})=(\frac{1}{2})^3+3(\frac{1}{2})^2+3(\frac{1}{2})+1\) ≠ 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

Problem: If \(ax^3 + bx^2 +x -6\) has x+2 as a factor and leaves reminder 4 when divided by x-2, find the values of a and b.

Solve:

To solve this problem, we can use the Factor Theorem and the Remainder Theorem. Given that the polynomial is divided by (x + 2) and leaves a remainder of 4 when divided by (x - 2), we can set up the following equations:

Using the Factor Theorem for (x + 2) as a factor: When (x + 2) is a factor of the polynomial, it means that if we substitute x = -2 into the polynomial, the result should be 0. So, we have:

\(a(-2)^3 + b(-2)^2 + (-2) - 6\) =0
\(\Rightarrow\) \(0 -8a + 4b - 2 - 6\) = 0
\(\therefore\) \(b = 2a+2\)

Using the Remainder Theorem for (x - 2) with a remainder of 4: When the polynomial is divided by (x - 2) and leaves a remainder of 4, it means that if we substitute x = 2 into the polynomial, the result should be 4. So, we have:

\(a(2)^3 + b(2)^2 + 2 - 6 =4\)
\(\Rightarrow\) \( 8a + 4b - 4 =4\)
\(\Rightarrow\) \( 8a + 4b = 8\)
\(\Rightarrow\) \(4b = 4(2-2a)\)
\(\therefore\) \(b = 2-2a\)

\( \therefore\) \(2a+2\)=\( 2-2a\)
\(\Rightarrow\) a=0

\(\therefore\) b=2-2*0=0