A theorem establishing the relationship between factors and zeros of a polynomial is a factor theorem. It is used when factoring the polynomials completely. If an algebraic expression is written as the product of algebraic expressions, then each of these expressions is called the factors of the given algebraic expression. A polynomial \(q(x)\) is said to divide a polynomial \(p(x)\) completely if the remainder is zero. If \(q(x)\) divides \(p(x)\), we say that \(p(x)\) is divisible by \(q(x)\) or, \(q(x)\) is a factor of \(p(x)\).

In this article, we will study a theorem that will enable us to determine whether polynomial \(q(x)\) is a factor of \(p(x)\) or not without performing actual division.

Factor Theorem

We have studied that a polynomial \(q(x)\) is said to divide a polynomial \(p(x)\) completely if the remainder is zero. If \(q(x)\) divides \(p(x)\), we say that \(p(x)\) is divisible by \(q(x)\) or, \(q(x)\) is a factor of \(p(x)\). In this article, we will study a theorem that will enable us to determine whether polynomial \(q(x)\) is a factor of \(p(x)\) or not without performing actual division.

This theorem is known as the factor theorem, which is stated and proved below.

Theorem: Let \(p(x)\) be a polynomial of degree greater than or equal to \(1\) and \(a\) be a real number such that \(p(a) = 0\), then \((x – a)\) is a factor of \(p(x)\).
Proof: Let \(p(x)\) be a polynomial of degree greater than or equal to \(1\) and \(a\) be a real number such that \(p(a) = 0\), then we have to show that \((x – a)\) is a factor of \(p(x)\).
Let \(q(x)\) be the quotient when \(p(x)\) is divided by \((x – a)\).
By remainder theorem, \(p(x)\) when divided by \((x – a)\) gives remainder equal to \(p(a)\).
Therefore, \(p(x) = (x – a)q(x) + p(a)\)
Putting \(p(a) = 0\), we get:
\(p(x) = (x – a) q(x)\)
\(\Rightarrow (x – a)\) is a factor of \(p(x)\)
Conversely, let \((x-a)\) be a factor of \(p(x)\). Then, we have to prove that \(p(a) = 0\)
Now, \((x – a)\) is a factor of \(p(x)\)
\(\Rightarrow p(x)\) when divided by \((x-a)\) gives remainder zero. But, by the remainder theorem, \(p(x)\) when divided by \((x – a)\) gives the remainder equal to \(p(a)\).
Therefore, \(p(a) = 0\).

Factor Theorem: Polynomials

An algebraic expression that consists of variables with exponents as whole numbers, coefficients, and constants combined using basic mathematical operations like addition, subtraction, and multiplication is called a polynomial.

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Different Types of Polynomials

Based on the number of terms in an expression, we can classify the polynomial as monomial, binomial, trinomial.
Monomial: A monomial is a polynomial that consists of only one term.
Example: \(4x,\,3y,\,x^2,\,y^3,\,3a^4\), etc.
Binomial: A binomial is a polynomial that consists of only two terms.
Example: \(x+1,\,x^2 -1,\,y^3+4,\,a+3,\,x^2+x\), etc.
Trinomial: A trinomial is an algebraic expression that consists of three terms.
Example: \(x^2 + x + 1,\,x^2 + y^2 + 2,\,y – 3x + 2\), etc.

Factor Theorem: Division of Algebraic Expressions

If \(x\) is a variable and \(m,\,n\) are positive integers such that \(m > n\) then \((x^m ÷ x^n ) = x^{m-n}\)

Division of Monomial by a Monomial

Rule: \(\text{Quotient of two monomials} = (\text{quotient}\,\text{of}\,\text{their}\,\text{numerical}\,\text{coefficients}) \times (\text{quotients}\,\text{of}\,\text{their}\,\text{variables})\)
Example: Divide \(8x^2 y^3\) by \(- 2xy\)
Solution: We have \(\frac{ {8 {x^2} {y^3}}}{ { – 2xy}}\)
We know that, \(\frac{ { {a^m}}}{ { {a^n}}} = {a^ {\left({m – n} \right)}}\)
So, \(\frac{{8{x^2}{y^3}}}{{ – 2xy}} = \left({\frac{8}{{ – 2}}} \right){x^{(2 – 1)}}{y^{(3 – 1)}} = – 4x{y^2}\)

Division of Polynomial by a Monomial

For dividing a polynomial by a monomial in the same variable, we perform the below steps:

1. Obtain the polynomial (dividend) and the monomial (divisor).
2. Arrange the terms of the dividend in descending order of their degrees.
3. Divide each term of a polynomial by the given monomial, using the rules of division of monomial by a monomial.

Example: Divide \(24x^3 y + 20x^2 y^2 – 4xy\) by \(2xy\)
Solution: We have, \(\frac{{24{x^3}y + 20{x^2}{y^2} – 4xy}}{{2xy}} = \frac{{24{x^3}y}}{{2xy}} + \frac{{20{x^2}{y^2}}}{{2xy}} – \frac{{4xy}}{{2xy}}\)
\(= 12x^2 + 10xy – 2\)

Division of Polynomial by a Binomial

For dividing a polynomial by a binomial, we will follow the following steps:

1. Arrange the terms of the dividend and divisor in descending order of their degrees.
2. Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
3. Multiply the divisor by the first term of the quotient and subtract the result from the dividend to obtain the remainder.
4. Consider the remainder (If any) as a new dividend and repeat step (ii) to obtain the second term of the quotient.
5. Repeat the above process till we obtain a remainder which is either zero or a polynomial of degree less than that of the divisor.

Example: Divide \(2x^4 + 8x^3 + 7x^2 + 4x + 3\) by \((x + 3)\)
Solution:

Division of Polynomial by a Polynomial

We may proceed according to the steps given below:

1. Arrange the terms of the dividend and divisor in descending order of their degrees.
2. Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
3. Multiply all the terms of the divisor by the first term of the quotient and subtract the result from the dividend.
4. Consider the remainder (if any) as a new dividend and proceed as before.
5. Repeat this process till we obtain a remainder which is either \(0\) or a polynomial of degree less than that of the divisor.

Example: Divide \(2x^2 + 3x + 1\) by \((x+1)\)
Solution:

Therefore, quotient \(= (2x + 1)\) and remainder \(= 0\).

Zeros of a Polynomial

The zeros of a polynomial \(p(x)\) are all the \(x-\)values that make the polynomial equal to zero.
Consider the polynomial \(f(x) = x^3 – 6x^2 + 11x – 6\)
The value of this polynomial at \(x = 1\) is \(f(1) = 1 – 6 + 11 – 6 = 0\)
So, we can say that \(1\) is the zero or root of the polynomial \(f(x)\).

Remainder Theorem

While dividing polynomials, we have noticed that when a polynomial \(f(x)\) is divided by a polynomial \(g(x)\) then the degree of the remainder is either zero or less than that of the divisor. It follows from this fact that when a polynomial \(f(x)\) is divided by a linear polynomial, then the degree of the remainder must be zero, i.e., it must be a constant, which may also be zero.

The remainder is \(5\).
Equate \(x – 1\) to \(0\).
\( \Rightarrow x – 1 = 0\)
\( \Rightarrow x = 1\)
Let us now compute \(p(1)\) i.e., the value of \(p(x)\) when \(x\) is replaced by \(1\).
We have, \(p(1) = 1 – 3 + 2 + 5 = 5\)
Thus, we find that the remainder when \(p(x)\) is divided by \(x – 1\) is equal to \(p(1)\) i.e., the value of \(p(x)\) at \(x = 1\).
The above result is stated as a theorem known as the remainder theorem.

Theorem: Let \(p(x)\) be any polynomial of degree greater than or equal to one, and \(a\) be any real number. If \(p(x)\) is divided by \((x – a)\), then the remainder is equal to \(p(a)\).
Proof: Let \(q(x)\) be the quotient and \(r(x)\) be the remainder when \(p(x)\) is divided by \((x – a)\). Then,
\(p(x) = (x – a) q(x) + r(x)\) ……..(i)
Where \(r(x) = 0\) or degree \(r(x) < \rm{degree} (x – a)\). But, \((x-a)\) is a polynomial of degree \(1\), and a polynomial of degree less than \(1\) is a constant. Therefore, either \(r(x) = 0\) or, \(r(x) = \rm{constant}\).
Let \(r(x) = r\). Then,
\(p(x) = (x – a)q(x) + r\) ………(ii)
Putting \(x = a\) in (ii), we get
\(p(a) = (a – a)q(a) + r\)
\(\Rightarrow p(a) = 0 \times q(a) + r \)
\(\Rightarrow p(a) = 0 + r\)
\(\Rightarrow p(a) = r\)
This shows that the remainder is \(p(a)\) when \(p(x)\) is divided by \((x – a)\).

Practice Exam Questions

Factorization of Polynomials using Factor Theorem

We have learned about the factorization of polynomials of the form \(ax^2 + bx + c\) by splitting the middle term. Now, we will use the factor theorem for the factorization of polynomials with integral coefficients. We will use the following theorems for finding the rational roots of a polynomial with integral coefficients.

Theorem 1: Integral root theorem: If \(f(x)\) is a polynomial with integral coefficients and leading coefficient \(1\), then any integer root of \(f(x)\) is a factor of the constant term.

Theorem 2: Rational root Theorem: Let \(f(x) = a_n x^n + a_{n-1} x^{n-1} +…….+ a_1 x + a_0\), \(a_n ≠ 0\) be a polynomial with integer coefficients and \(\frac{b}{c}\) be a rational fraction in the lowest terms. Then, \(\frac{b}{c}\) is a root \(f(x)\), if \(b\) is a factor of the constant term \(a_0\) and \(c\) is a factor of the leading coefficient \(a_n\).

Solved Examples

Q.1. Show that \((x – 3)\) is a factor of the polynomial \(x^3 – 3x^2 + 4x – 12\).
Ans: Let \(p(x) =x^3 – 3x^2 + 4x – 12\) be the given polynomial. By factor theorem, \((x-a)\) is a factor of polynomial \(p(x)\) if \(p(a) = 0\). Therefore, to prove that \((x – 3)\) is a factor of \(p(x)\), it is sufficient to show that \(p(3) = 0\).
Now, \(p(x) = x^3 – 3x^2 + 4x – 12\)
\(\Rightarrow p(3) = 3^3 – 3 \times 3^2 + 4 \times 3 – 12 = 27 – 27 + 12 – 12 = 0\)
Hence, \((x – 3)\) is a factor of \(p(x) = x^3 – 3x^2 + 4x – 12\).

Q.2. Examine whether \(x + 2\) is a factor of \(x^3 + 3x^2 + 5x + 6\).
Ans: The zero of \(x + 2\) is \(- 2\).
\(p(x) = x^3 + 3x^2 + 5x + 6\)
Then, \(p(- 2) = (- 2)^3 + 3 \times 2^2 + 5 \times (-2) + 6 = – 8 + 12 – 10 + 6 = 0\)
So, by factor theorem, \((x+2)\) is a factor of \(x^3 – 3x^2 + 4x – 12\).

Q.3. Find the value of \(k\), if \(x – 1\) is a factor of \(4x^3 + 3x^2 – 4x + k\).
Ans: As \(x – 1\) is a factor of \(p(x) = 4x^3 + 3x^2 – 4x + k\), \(p(1) = 0\).
Now, \(p(1) = 4(1)^3 + 3(1)^2 – 4(1) + k\)
\(\Rightarrow 4 + 3 – 4 + k = 0\)
\(\Rightarrow k = – 3\)

Q.4. Find the value of \(a\), if \((x – a)\) is a factor of \(x^3 – a^2 x + x + 2\).
Ans: Let \(p(x) = x^3 – a^2 x + x + 2\)
The zero of \(x – a\) is \(a\).
\(\Rightarrow p(a) = 0\)
Now, \(p(a) = a^3 – a^2 \times a + a + 2 = 0\)
\(\Rightarrow a + 2 = 0\)
\(\Rightarrow a = – 2\)
Therefore, \((x – a)\) is a factor of the given polynomial if \(a = – 2\).

Q.5. Determine the value of a for which the polynomial \(2x^4 – ax^3 + 4x^2 + 2x + 1\) is divisible by \(1 – 2x\).
Ans: Let \(p(x) = 2x^4 – ax^3 + 4x^2 + 2x + 1\). If the polynomial \(p(x)\) is divisible by \((1 – 2x)\), then \((1 – 2x)\) is a factor of \(p(x)\).
Therefore, \(p\left({\frac{1}{2}} \right) = 0\)
\( \Rightarrow 2{\left( {\frac{1}{2}} \right)^4} – a{\left( {\frac{1}{2}} \right)^3} + 4{\left( {\frac{1}{2}} \right)^2} + 2\left( {\frac{1}{2}} \right) + 1 = 0\)
\( \Rightarrow \frac{2}{ {16}} – \frac{a}{8} + \frac{4}{4} + \frac{2}{2} + 1 = 0\)
\( \Rightarrow \frac{1}{ {8}} – \frac{a}{8} + 1 + 1 + 1 = 0\)
\( \Rightarrow \frac{{25}}{8} = \frac{a}{8}\)
\( \Rightarrow a = 25\)
Hence, the given polynomial will be divisible by \(1 – 2x\), if \(a = 25\).

Summary

This article explains the division of polynomials, remainder theorem, factor theorem, and factorization of polynomials using factor theorem. Also, we solved some example problems using the factor theorem, which will help you understand the concept better.

FAQs

Q.1. What does the factor theorem tell us?
Ans: Factor theorem tells us that, let \(p(x)\) be a polynomial of degree greater than or equal to \(1\) and \(a\) be a real number such that \(p(a) = 0\), then \((x – a)\) is a factor of \(p(x)\).
Example: If \((x – 3)\) is a factor of \(p(x) = x^3 – 3x^2 + 4x – 12\). Then \(p(3) = 0\).

Q.2. What is the factor theorem? Explain with example.
Ans: Factor theorem states that, let \(p(x)\) be a polynomial of degree greater than or equal to \(1\) and \(a\) be a real number such that \(p(a) = 0\), then \((x-a)\) is a factor of \(p(x)\).
Example: If \((x – 3)\) is a factor of \(p(x) = x^3 – 3x^2 + 4x – 12\). Then \(p(3) = 0\).

Q.3. What is the factor theorem formula?
Ans: There is no particular formula for the factor theorem. The theorem states that, let \(p(x)\) be a polynomial of degree greater than or equal to \(1\) and \(a\) be a real number such that \(p(a) = 0\), then \((x – a)\) is a factor of \(p(x)\).

Q.4. State the remainder theorem.
Ans: Let \(p(x)\) be any polynomial of degree greater than or equal to one, and \(a\) be any real number. If \(p(x)\) is divided by \((x – a)\), then the remainder is equal to \(p(a)\).

Q.5. Who discovered the factor theorem?
Ans: Polish school students learn about the Factor Theorem under the name “Twierdzenie Bézout” (Bézout Theorem) after Etienne Bézout discovered the remainder theorem.

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