Factorization of Polynomials | Mathematics (Maths) Class 9 PDF Download

Factor theorem

Factor theorem: Let p(x) be a polynomial of degree n ≥ 1 and let a be any real number. Then the following are equivalent.

• If p(a) = 0, then (x - a) is a factor of p(x).
• If (x - a) is a factor of p(x), then p(a) = 0.

Factor Theorem 

Proof

(i) Assume p(a) = 0. Consider division of p(x) by (x - a).

When p(x) is divided by (x - a) the remainder is p(a) (Remainder Theorem).

Dividend = Divisor × Quotient + Remainder

Therefore, p(x) = (x - a)·g(x) + p(a).

Since p(a) = 0, p(x) = (x - a)·g(x).

Hence, (x - a) is a factor of p(x).

(ii) Assume (x - a) is a factor of p(x). Then p(x) = (x - a)·g(x) for some polynomial g(x).

Substitute x = a into p(x) = (x - a)·g(x).

We get p(a) = (a - a)·g(a) = 0.

Thus, if (x - a) is a factor of p(x) then p(a) = 0.

Example 1

Examine whether x + 1 is a factor of p(x) = x³ + 3x² + 5x + 6.

We check if p(-1) = 0, because x + 1 = x - (-1).

p(x) = x³ + 3x² + 5x + 6

p(-1) = (-1)³ + 3(-1)² + 5(-1) + 6

p(-1) = -1 + 3 - 5 + 6

p(-1) = 3 ≠ 0

Therefore x + 1 is not a factor of p(x).

Example 2

Find the value of a if (x - a) is a factor of x³ - a²x + x + 2.

Let p(x) = x³ - a²x + x + 2.

If (x - a) is a factor, then p(a) = 0.

p(a) = a³ - a²·a + a + 2

p(a) = a³ - a³ + a + 2

p(a) = a + 2 = 0

So a = -2.

Hence (x - a) is a factor of the given polynomial when a = -2.

Factorisation of a quadratic polynomial

A quadratic polynomial has the general form ax² + bx + c with a ≠ 0. It can be factorised by different methods.

• By splitting the middle term
• By using factor theorem (finding integer roots)

(i) By splitting the middle term

Assume ax² + bx + c = (px + q)(rx + s).

Expanding gives ax² + bx + c = prx² + (ps + qr)x + qs.

Comparing coefficients, a = pr, b = ps + qr and c = qs.

Thus b can be written as sum of two numbers whose product is ac. We split the middle term bx into two terms whose coefficients multiply to ac, then factor by grouping.

Example

Factorise 2x² + 7x + 3 by splitting the middle term.

Compare with ax² + bx + c: a = 2, b = 7, c = 3.

Compute ac = 2 × 3 = 6.

Find two integers whose product is 6 and sum is b = 7. The pair 1 and 6 works.

Write 7x as (1 + 6)x:

2x² + 7x + 3 = 2x² + x + 6x + 3

Group terms: = x(2x + 1) + 3(2x + 1)

Factor common term: = (2x + 1)(x + 3)

(ii) By using factor theorem

When the coefficient of x² is 1, we can find integer zeros by checking divisors of the constant term. If x = r makes the polynomial zero then (x - r) is a factor. If the coefficient of x² is not 1, first write the polynomial as a·g(x) where g(x) has leading coefficient 1.

Example

Factorise x² - 5x + 6 using the factor theorem.

Let f(x) = x² - 5x + 6.

Factors of the constant term 6 are ±1, ±2, ±3, ±6. Test x = 2 and x = 3.

f(2) = 2² - 5·2 + 6 = 4 - 10 + 6 = 0

f(3) = 3² - 5·3 + 6 = 9 - 15 + 6 = 0

Therefore (x - 2) and (x - 3) are factors and

x² - 5x + 6 = (x - 2)(x - 3).

Factorisation of a cubic polynomial

We use the following steps to factorise a cubic polynomial,
Step I: Write the given cubic polynomial, p(x) = ax³+ bx² + cx + d as,

i.e., first make the coefficient of x= equal to 1 if it is not 1, then find the constant term.

Step II: Find all the possible factors of constant term (d/a) of g(X).

Step III: Check at which factor of constant term, p(x) is zero and get one factor of p(x), (i.e., x − α)
Step IV: Write p(x) as the product of this factor (x − α) and a quadratic polynomial i.e., p(x) = (x − α) (a₁x² + b₁x + c₁)
Step V: Apply splitting method of middle term or factor theorem in quadratic polynomial to get the other two factors

Example 1

Factorise x³ + 13x² + 32x + 20 using the factor theorem.

Let p(x) = x³ + 13x² + 32x + 20.

Constant term = 20 and leading coefficient = 1, so possible integer roots are ±1, ±2, ±4, ±5, ±10, ±20.

Evaluate p(-1):

p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20

p(-1) = -1 + 13 - 32 + 20

p(-1) = 0

Thus (x + 1) is a factor.

Write p(x) grouping terms to factor (x + 1) by common factor:

x³ + 13x² + 32x + 20 = x²(x + 1) + 12x(x + 1) + 20(x + 1)

= (x + 1)(x² + 12x + 20)

Now factor the quadratic x² + 12x + 20 by splitting the middle term:

x² + 12x + 20 = x² + 10x + 2x + 20

= x(x + 10) + 2(x + 10)

= (x + 2)(x + 10)

Hence:

x³ + 13x² + 32x + 20 = (x + 1)(x + 2)(x + 10)

Example 2

If p(y) = y³ - 4y² + y + 6 then show that p(3) = 0 and hence factorise p(y).

Given p(y) = y³ - 4y² + y + 6.

Evaluate p(3):

p(3) = 3³ - 4·3² + 3 + 6

p(3) = 27 - 36 + 3 + 6 = 0

So (y - 3) is a factor.

Divide p(y) by (y - 3) or factor by grouping to obtain the quadratic factor:

p(y) = (y - 3)(y² - y - 2)

Factor the quadratic y² - y - 2 by splitting the middle term (-2 and +1):

y² - y - 2 = y² - 2y + y - 2

= y(y - 2) + 1(y - 2)

= (y - 2)(y + 1)

Therefore:

p(y) = (y - 3)(y + 1)(y - 2)

Summary

The factor theorem links roots and linear factors: p(a) = 0 ⇔ (x - a) is a factor of p(x). For quadratics, use splitting of the middle term or test integer roots when leading coefficient is 1. For cubics, list possible rational roots, test them using the factor theorem, extract a linear factor, and then factor the resulting quadratic. These procedures are essential tools for simplifying polynomials and solving polynomial equations.