P(x) = (x - a) Q(x). In other words, (x - a) is said to be the factor of polynomial P(x); if P(a) = 0. This theorem tell

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Factor Theorem

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Zero Factor Theorem

Polynomial is a very important topic in mathematics. Polynomials are used everywhere in maths. Polynomials are an expression of

Factor Theorem Proof

the form -

Factor Theorem Examples

Where, n is a positive integer and

.

There are two important theorems that play a vital role while dealing with polynomials. One is remainder theorem and other is

Related Concepts

factor theorem. Factor theorem is derived from the remainder theorem. Let us first learn about the remainder theorem, according

Polynomial Factor Theorem

If P(x) be a polynomial and it is divided by another polynomial (x - a), then the quotient is represented by Q(x) and the remainder

Prime Factorization Theorem

to which is given by R(x). Then remainder theorem says that P(x) = (x - a) Q(x) + R(x)

Factor Apothem Theorem

Factor theorem is a special case of remainder theorem. Factor theorem states that if a polynomial P(x) is evenly divided by another

Baye's Theorem

polynomial (x - a), then it leaves no remainder; i.e. R(x) = 0 and hence (x - a) is said to be the factor of P(x). According to factor Binomial Theorems

theorem -

Bisector Theorem

P(x) = (x - a) Q(x)

Calculus Theorems

In other words, (x - a) is said to be the factor of polynomial P(x); if P(a) = 0.

Related Formulas

This theorem tells us that there is a relation between factors of the polynomial and zeros of the polynomial. This theorem makes us find the roots of the polynomials and also helps us to solve the polynomials of higher degree.

Equation by Factoring Examples of Pythagorean Theorem Problems

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Factor Theorem Definition

Factor by Grouping Factor Tree

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The factor theorem is a theorem commonly applied to factorizing and finding the roots of polynomial equations. If P(x), a polynomial in x is divided by x - a and the remainder is zero, then (x - a) is a factor of P(x).

Related Worksheets

P(x) = (x - a) Q(x)

Factoring Worksheets

Zero Factor Theorem

Factors Worksheet

Zero factor theorem is the inverse case of the factor theorem. It is used to find the factors of the polynomial equations. This

Central Limit Theorem Worksheet Converse of the Pythagorean Theorem Worksheet

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theorem states that, if x - r is the factor of P(x), then r is a zero of the polynomial P(x).

Solved Example Question: Find the roots of the x 2 + x - 6 = 0 Solution: Given polynomial equation is, x 2 + x - 6 = 0 x 2 + x - 6 = 0 => x 2 + 3x - 2x - 6 = 0 => x(x + 3) - 2(x - 3) = 0 => (x - 2)(x + 3) = 0 Product of two binomials is zero. So, either x - 2 = 0 or x + 3 = 0 => x = 2 or x = -3 Therefore, roots of the given polynomial equation are 2, -3.

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Factor Theorem Proof

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The polynomial P(x) is exactly divisible by the x - a iff the value of the polynomial is zero for x = a. Proof: Since P(x) is divided by x - a P(x) = (x - a) . Q(x) + P(a) ( by remainder theorem ) (Dividend = Divisor x quotient + Remainder Division Algorithm) But, P(a) = 0 is given. => P(x) = (x - a) . Q(x) => (x - a) is a factor of P(x). Converse: If x - a is a factor of P(x), then P(a) = 0. P(x) = (x - a) . Q(x) + P(a) If (x - a) is a factor, then the remainder should be zero (x - a divides P(x) exactly) P(a) = 0

Factor Theorem Examples

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Given below are some of the examples on factor theorem.

Solved Examples Question 1: Determine whether x - 2 is a factor of x 2 - 7x + 10. Solution: Let P(x) = x 2 - 7x + 10 By the factor theorem, x - 2 is the factor of P(x) if P(2) = 0. Find the value of P(2). P(x) = x 2 - 7x + 10 P(2) = 2 2 - 7 * 2 + 10 (Substitute x = 2) = 4 - 14 + 10 = 0 Since P(2) = 0 => x - 2 is a factor of P(x).

Question 2: Determine whether x - 3 is a factor of x 3 - 3x 2 + 4x - 12. Solution: Let P(x) = x 3 - 3x 2 + 4x - 12 By the factor theorem, x - 3 is the factor of P(x) if P(3) = 0. Find the value of P(3). P(x) = x 3 - 3x 2 + 4x - 12 P(3) = 3 3 - 3 * 3 2 + 4 * 3 - 12 (Substitute x = 3) = 27 - 27 + 12 - 12 = 0 => x - 3 is a factor of P(x).

Question 3: Show that x + 1 is a factor of 2x 3 + 5x 2 - 9x - 12. Solution: Let P(x) = 2x 3 + 5x 2 - 9x - 12 By the factor theorem, x - (-1) is the factor of P(x) if P(-1) = 0. Find the value of P(-1). P(x) = 2x 3 + 5x 2 - 9x - 12 P(x) = 2(-1)3 + 5(-1)2 - 9(-1) - 12 = -2 + 5 + 9 - 12 = 0 => x + 1 is a factor of P(x).

Question 4: Find the value of a so that x 4 + 2x 3 - ax 2 + x - 2 has (x + 2) as its factor. Solution: Let P(x) = x 4 + 2x 3 - ax 2 + x - 2 Since x + 2 is a factor of P(x), so P(-2) = 0 => P(-2) = (-2)4 + 2(-2)3 - a(-2)2 + (-2) - 2 = 0 (substitute x = -2) => 16 - 16 - 4a - 2 - 2 = 0 => -4a - 4 = 0 => -4a = 4 => a = -1 So, the value of a is -1.

Question 5: Factorize 2x 2 + x - 3 using factor theorem. Solution: Let P(x) = 2x 2 + x - 3 The factors of the constant 3 are +1, -1, +3, -3. Put x = 1 in P(x) => P(1) = 2 + 1 - 3 = 0 => x - 1 is one of a factor of P(x). Now, divide 2x 2 + x - 3 by x - 1 to get the other factor. 2x 2 + x - 3 = (2x + 3)(x - 1)

Question 6: If (x - 2) and (x - 3) are factors of x 3 + ax 2 + bx + 12, find a and b. Solution: Let P(x) = x 3 + ax 2 + bx + 12 Step 1: Case 1: Since x - 2 is a factor of P(x), so P(2) = 0 => 2 3 + a * 2 2 + b * 2 + 12 = 0 => 8 + 4a + 2b + 12 = 0 => 4a + 2b + 20 = 0 2a + b + 10 = 0 .............................(i) Case 2: Since x - 3 is a factor of P(x), so P(3) = 0 => 3 3 + a * 3 2 + b * 3 + 12 = 0 => 27 + 9a + 3b + 12 = 0 => 9a + 3b + 39 = 0 3a + b + 13 = 0 ................................(ii) Step 2: Subtracting equation (i) from (ii), we get, => 3a + b + 13 - (2a + b + 10) = 0 => 3a - 2a + b - b + 13 - 10 = 0 => a = -3 Substitute a = -3 in equation (1), 2a + b + 10 = 0 -6 + b = -10 b = -4

Zeros of Polynomials

Remainder Theorem

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