## Factorization Of Polynomials Using Factor Theorem

### Factor Theorem:

If p(x) is a polynomial of degree n 1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).

**Proof:** By the Remainder Theorem,

p(x) = (x – a) q(x) + p(a).

(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).

(ii) Since x – a is a factor of p(x),

p(x) = (x – a) g(x) for same polynomial g(x). In this case, p(a) = (a – a) g(a) = 0.

- Obtain the polynomial p(x).
- Obtain the constant term in p(x) and find its all possible factors. For example, in the polynomial

x^{4}+ x^{3}– 7x^{2}– x + 6 the constant term is 6 and its factors are ± 1, ± 2, ± 3, ± 6. - Take one of the factors, say a and replace x by it in the given polynomial. If the polynomial reduces to zero, then (x – a) is a factor of polynomial.
- Obtain the factors equal in no. to the degree of polynomial. Let these are (x–a), (x–b), (x–c.)…..
- Write p(x) = k (x–a) (x–b) (x–c) ….. where k is constant.
- Substitute any value of x other than a,b,c …… and find the value of k.

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## Factorization Of Polynomials Using Factor Theorem Example Problems With Solutions

**Example 1: **Factorize x^{2} +4 + 9 z^{2} + 4x – 6 xz – 12 z

**Solution:**

The presence of the three squares viz.x^{2}, (2)^{2}, and (3z)^{2} gives a clue that identity (vii) could be used. So we write.

A = x^{2} + (2)^{2} + (3z)^{2} + 4x – 6 xz – 12 z

We note that the last two of the product terms are negative and that both of these contain z. Hence we write A as

A = x^{2} + (2)^{2} + (–3z)^{2} + 2.2x – 2.x.(–3z) + 2.2 (– 3z)

= (x+2 – 3z)^{2}

= (x + 2 – 3z) (x + 2 – 3z)

**Example 2: ** Using factor theorem, factorize the polynomial x^{3} – 6x^{2} + 11 x – 6.

**Solution:**

Let f(x) = x^{3} – 6x^{2} + 11x – 6

The constant term in f(x) is equal to – 6 and factors of – 6 are ±1, ± 2, ± 3, ± 6.

Putting x = 1 in f(x), we have

f(1) = 1^{3} – 6 ×1^{2} + 11× 1– 6

= 1 – 6 + 11– 6 = 0

∴ (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 can not have more than three linear factors.

Let f(x) = k (x–1) (x– 2) (x – 3). Then,

x^{3}– 6x^{2} + 11x – 6 = k(x–1) (x– 2) (x– 3)

Putting x = 0 on both sides, we get

– 6 = k (0 – 1) (0 – 2) (0 – 3)

⇒ – 6 = – 6 k ⇒ k = 1

Putting k = 1 in f(x) = k (x– 1) (x– 2) (x–3), we get

f(x) = (x–1) (x– 2) (x – 3)

Hence, x^{3}–6x^{2} + 11x – 6 = (x– 1) (x – 2) (x–3)

**Example 3: **Using factor theorem, factorize the polynomial x^{4} + x^{3} – 7x^{2} – x + 6.

**Solution:**

Let f(x) = x^{4} + x^{3}– 7x^{2} –x + 6

the factors of constant term in f(x) are ±1, ±2, ±3 and ± 6

Now,

Since f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors

Thus, the factors of f (x) are (x–1), (x+1),

(x–2) and (x+3).

Let f(x) = k (x–1) (x+1) (x–2) (x + 3)

⇒ x^{4} + x^{3} – 7x^{2} – x + 6

= k (x–1) (x +1) (x – 2) (x + 3)

Putting x = 0 on both sides, we get

6 = k (–1) (1) (–2) (3) ⇒ 6 = 6 k ⇒ k = 1

Substituting k = 1 in (i), we get

x^{4} + x^{3} – 7x^{2} – x + 6 = (x–1) (x +1) (x–2) (x+3)

**Example 4: **Factorize, 2x^{4} + x^{3} – 14x^{2} – 19x – 6

**Solution:**

Let f(x) = 2x^{4} + x^{3} – 14x^{2} – 19x – 6 be the given polynomial. The factors of the constant term – 6 are ±1, ±2, ±3 and ±6, we have,

f(–1) = 2(–1)^{4} + (–1)^{3} – 14(–1)^{2} – 19(–1)– 6

= 2 – 1 – 14 + 19 – 6 = 21 – 21 = 0

and,

f(–2) = 2(–2)^{4} + (–2)^{3} – 14(–2)^{2} – 19(–2)– 6

= 32 – 8 – 56 + 38 – 6 = 0

So, x + 1 and x + 2 are factors of f(x).

⇒ (x + 1) (x + 2) is also a factor of f(x)

⇒ x^{2} + 3x + 2 is a factor of f(x)

Now, we divide

f(x) = 2x^{4} +x^{3} – 14x^{2}–19x – 6 by

x^{2} + 3x + 2 to get the other factors.

**Example 5: **Factorize, 9z^{3} – 27z^{2} – 100 z+ 300, if it is given that (3z+10) is a factor of it.

**Solution:**

Let us divide 9z^{3} – 27z^{2} – 100 z+ 300 by

3z + 10 to get the other factors

∴ 9z^{3} – 27z^{2} – 100 z+ 300

= (3z + 10) (3z^{2}–19z + 30)

= (3z + 10) (3z^{2}–10z – 9z + 30)

= (3z + 10) {(3z^{2}–10z) – (9z – 30)}

= (3z + 10) {z(3z–10) – 3(3z–10)}

= (3z + 10) (3z–10) (z–3)

Hence, 9z^{3}–27z^{2}–100z+ 300

= (3z + 10) (3z–10) (z–3)

**Example 6: **Simplify

\(\frac{4x-2}{{{x}^{2}}-x-2}+\frac{3}{2{{x}^{2}}-7x+6}-\frac{8x+3}{2{{x}^{2}}-x-3}\)

**Solution:**

**Example 7: **Establish the identity

\(\frac{6{{x}^{2}}+11x-8}{3x-2}=\left( 2x+5 \right)+\frac{2}{3x-2}\)

**Solution:**