# Application of Factor Theorem | Using Factor Theorem to find Factors of a Polynomial and Roots of Equation

In maths, the factor theorem will establish a relationship between the factors and zeros of a polynomial. It is a unique case consideration of the polynomial remainder theorem, Factor theorem is mainly used to factor the polynomials and to find the n-roots of the polynomials. It will be very helpful for analyzing the polynomial equations.

This theorem will be considered a special case for the polynomial remainder theorem. It will be frequently linked with the remainder theorem, therefore do not confuse both. On this page, we will discuss the 10th Grade Math Concept Application of factor theorem, how to use factor theorem, example problems, and so on.

### What is the Application of the Factor Theorem?

Factor theorem is usually used to factor and find the roots of polynomials. In a polynomial, the roots or zeros are equal to zero. Therefore, this factor theorem simply states that when f(k) = 0, then (x – k) is a factor of f(x).

### Application of Factor Theorem Examples

Example 1: Use the factor theorem to check whether the y + 1 is a factor of the polynomial 3y4 + y3 – y2 + 3y + 2 or not.

Solution:
Given that, the y + 1 is a factor.
Then, y + 1 = 0.
we get y = -1.
Now, substitute y = -1 in the given polynomial equation.
The polynomial equation is 3y4 + y3 – y2 + 3y + 2.
After substituting, 3(–1)4 + (–1)3 – (–1)2 +3(–1) + 2.
= 3(1) + (–1) – 1 – 3 + 2
= 3 -1 -1 -3 + 2
Adding all the positive terms and all the negative terms. we get,
5 – 5= 0
Therefore, we can say that y + 1 is a factor of 3y4 + y3 – y2 + 3y + 2.

Example 2: Check whether 2y + 1 is a factor of the polynomial 4y3 + 4y2 – y – 1 or not using the factor theorem.

Solution:
In the given question,
The factor is 2y+1.
The polynomial equation is 4y3 + 4y2 – y – 1
Let’s equate the given binomial to zero.
i.e., 2y + 1 = 0.
Therefore,  y = -1/2
Now substitute y = -1/2 in the given polynomial equation 4y3 + 4y2 – y – 1.
4( -1/2)3 + 4(-1/2)2 – (-1/2) – 1
= -1/2 + 1 + 1/2 – 1 = 0
So, the remainder is 0.
Thus 2y + 1 is a factor of the polynomial equation 4y3 + 4y2 – y – 1.

Example 3: Verify (2x+3) and (x+3) are the linear factors of the quadratic equation f(x) = 2x2 + 9x + 9.

Solution:
Given that,
To verify if (2x+3) and (x+3) are the linear factors of the quadratic equation f(x) = 2x2 + 9x + 9.
First, we will multiply the factors.
(2x + 3)(x + 3) = 2x2 + 3x + 6x + 9 = 2x2 + 9x + 9.
Hence, (2x+3) and (x+3) are the linear factors of the quadratic equation f(x) = 2x2 + 9x + 9.

Example 4: Find the factors of the quadratic equation x2 + x – 12 = 0 using the factoring quadratics method.

Solution:
In the given question, the quadratic equation is x2 + x – 12 = 0.
Now, we need to find out the factors of the given equation.
First, split the middle term of the quadratic equation x2 + x – 12 = 0 to determine its factors.
i.e., x2 + x – 12 = 0
x2 + 4x – 3x – 12 = 0
x(x + 4) – 3(x + 4) = 0
(x – 3)(x + 4) = 0
Hence the factors of x2 + x – 12 = 0 are (x – 3) and (x + 4).

Example 5: Find the cubic equation whose roots are 2, √3, and -√3.

Solution:
Given that, the roots are 2, √3, and -√3.
Now, we will find the cubic equation.
We know the quadratic equation whose roots are α, β, and γ, is
(x – α)(x – β)(x – γ) = 0
So, the required equation is (x – 2)(x – √3)(x – (-√3)) = 0
(x – 2)(x – √3)(x + √3) = 0
(x – 2)(x2 – 3) = 0
x3 – 2x2 – 3x + 6 = 0.
x2  – (1 – √3 + 1 + √3)x + (1 + √3)( 1 – √3) = 0
x2  – 2x + (1 – 3) = 0
x2  – 2x – 2 = 0.
Thus, the equation is x2 -2x-2.

Example 6: Factorize x2 -3x – 9.

Solution:
As given in the question,
The polynomial equation is x2 – 3x – 9= 0
Now we apply the quadratic formula,
x = −b±√b2-4ac/2a
Substitute the values, we get
x= −(−3)±√(-3)2-4(1)(-9)/2(1)
= 3±√9+36/2
= 3±√45/2
= 3±3√5/2
Therefore, x2 – 3x – 9 = (x – 3+3√5/2)(x – 3−3√5/2)

Example 7: Consider an quadratic equation x2+ 5x + 4 = 0. It is in the form of ax2 + bx + c = 0. The value of a is 1, b is 5, c is 4. Find the factors of the quadratic equation?

Solution:
As given in the question,
The quadratic equation is x2+ 5x + 4 = 0
First, we will substitute the values of a, b, and c in the quadratic formula, we get
x=−b±√b2−4ac/2a
x=−5±√52−4(1)(4) /2(1)
x =−5±√25−16/2
x =−5±√9/2
x =-5±3/2 which means −5+3/2 and −5−3/2
So, x= −2/2 and x =− 8/2
x= −1 and −4
Therefore, the factors of the given equations are (x + 1) and (x + 4).

Example 8: For a curve that crosses the x-axis at 3 points, of which one is at 2. What is the factor of 2x3−x2−7x+2?

Solution:
Given that, the polynomial expression is 2x3−x2−7x+2
Now, we need to find the factors of the given polynomial.
First, find the degree of the polynomial.
The polynomial for the equation degree is 3 and could be all easy to solve.
Now, substitute x value as ‘2’ in the given polynomial, then we get
f(2) = 2(2)3−(2)2−7(2)+2
Now, this will be, 16−4−14+2 =12 (-14) +2
So, the value is -2 + 2 = 0
The f(2) =0 will have a factor and a root.
Therefore, the (x-2) should be a factor of 2x3−x2−7x+2.

Example 9: Using the factor theorem, factorize the polynomial x3 – 6x2 + 11 x – 6.

Solution:
Given that, the polynomial equation is f(x) = x3 – 6x2 + 11x – 6
The constant term in f(x) is equal to – 6 and the factors of – 6 are ±1, ± 2, ± 3, ± 6.
Initially, Put x = 1 in f(x), we have
f(1) = 13 – 6(1)2 + 11(1)– 6
= 1 – 6 + 11– 6 = 0
Therefore, the (x– 1) is a factor of f(x).
Similarly, x – 2 and x – 3 are factors of f(x).
Since f(x) is a polynomial of degree 3.
So, it will not have more than three linear factors.
Let f(x) = k (x–1) (x– 2) (x – 3). Then,
x3 – 6x2 + 11x – 6 = k(x–1) (x– 2) (x– 3)
Put x = 0 on both sides, we get
– 6 = k (0 – 1) (0 – 2) (0 – 3)
= – 6 = – 6 k ⇒ k = 1
Now, put k = 1 in f(x) = k (x– 1) (x– 2) (x–3), we get
f(x) = (x–1) (x– 2) (x – 3)
Hence, after factorize the equation x3 – 6x2 + 11x – 6. The factors are (x– 1) (x – 2) (x–3).

### FAQ’s on Application of Factor Theorem

1. Does the factor theorem work for quadratic equations?

We most commonly see it when dealing with quadratics. Think about how many problems have had you simplify something like x2+4x+3 into (x+1)(x+3). We use this process when we solve quadratics because of the factor theorem which states: “A polynomial, f(x), has a factor, (x−k), if and only if f(k)=0”.

2. What is the factor theorem in a quadratic equation?

It is a theorem that links factors and zeros of the polynomial. According to the factor theorem, if f(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number, then, (x-a) is a factor of f(x), if f(a)=0. Also, we say that if (x-a) is a factor of polynomial f(x), then f(a) is 0.

3. Why is the factor theorem useful?
We can use the factor theorem to completely factor a polynomial into the product of ‘n’ factors. Once the polynomial has been completely factored then we can easily determine the zeros of the polynomial.

4. What is the importance of the remainder theorem and factor theorem?

The remainder theorem and factor theorem are very handy tools. They tell us that we can find factors of a polynomial without using long division, synthetic division, or other traditional methods of factoring. Using the remainder theorem and the factor theorem is somewhat of a trial and error method.

5. What is the difference between the factor theorem and the remainder theorem?

Basically, the remainder theorem links the remainder of division by a binomial with the value of a function at a point, while the factor theorem links the factors of a polynomial to its zeros.

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