Exams Exam Factor Theorem: Definition and Examples

Written By Chaitanyesh Rudracharya
Last Modified 25-01-2023

Factor Theorem: Definition and Examples

Factor Theorem: In simple words, factor theorem is a theorem that goes on to establish a link between factors and zeros of a polynomial. It is used to factor a polynomial. Factorisation is expressing a mathematical expression or object into its products. For example, 2*4 is a factorisation of integer 8.

In algebra, when an expression can be expressed as its products, then each of them is called a factor of the expression.

Factor Theorem: Proof

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)

⇒(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)

⇒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.

Also Read: Know more on factors

What Are Polynomials?

An expression which has variables with exponents as whole numbers. It also has coefficients and constants. These are combined with the help of basic mathematical operations like addition, subtraction, and multiplication.

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)<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)=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

⇒p(a)=0×q(a)+r

⇒p(a)=0+r

⇒p(a)=r

This shows that the remainder is p(a) when p(x) is divided by (x–a).

Also Read: What are common factors?

Solved Examples

Let us look at some of the solved examples below:

Q.1. Show that (x–3) is a factor of the polynomial x3–3×2+4x–12.

Ans: Let p(x)=x3–3×2+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)=x3–3×2+4x–12

⇒p(3)=33–3×32+4×3–12=27–27+12–12=0

Hence, (x–3) is a factor of p(x)=x3–3×2+4x–12.

Q.2. Examine whether x+2 is a factor of x3+3×2+5x+6.

Ans: The zero of x+2 is −2.

p(x)=x3+3×2+5x+6

Then, p(−2)=(−2)3+3×22+5×(−2)+6=–8+12–10+6=0

So, by factor theorem, (x+2) is a factor of x3–3×2+4x–12.

Q.3. Find the value of k, if x–1 is a factor of 4×3+3×2–4x+k.

Ans: As x–1 is a factor of p(x)=4×3+3×2–4x+k, p(1)=0.

Now, p(1)=4(1)3+3(1)2–4(1)+k

⇒4+3–4+k=0

⇒k=–3

Q.4. Find the value of a, if (x–a) is a factor of x3–a2x+x+2.

Ans: Let p(x)=x3–a2x+x+2

The zero of x–a is a.

⇒p(a)=0

Now, p(a)=a3–a2×a+a+2=0

⇒a+2=0

⇒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 2×4–ax3+4×2+2x+1 is divisible by 1–2x.

Ans: Let p(x)=2×4–ax3+4×2+2x+1. If the polynomial p(x) is divisible by (1–2x), then (1–2x) is a factor of p(x).

Therefore, p(12)=0

⇒2(12)4–a(12)3+4(12)2+2(12)+1=0

⇒216–a8+44+22+1=0

⇒18–a8+1+1+1=0

⇒258=a8

⇒a=25

Hence, the given polynomial will be divisible by 1–2x, if a=25.

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