Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics PDF Download

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), then p(a) = 0 and

(ii) p(a) = 0 then x – a is a factor of p(x).

Ex. Use the factor theorem to determine whether x – 1 is a factor of

(a) x³ + 8x² – 7x – 2

(b) 2√2 x³ + 5√2 x² – 7√2

Sol. (a) Let p(x) = x³ + 8x² – 7x – 2

By using factor theorem, (x – 1) is a factor of p(x) only when p(1) = 0

p(1) = (1)³ + 8(1)² – 7(1) – 2 = 1 + 8 – 7 – 2 = 9 – 9 = 0

Hence (x – 1) is a factor of p(x).

(b) Let p(x) = 2√2x³ + 5√2x² – 7√2

By using factor theorem, (x – 1) is a factor of p(x) only when p(1) = 0

p(1) = 2√2 (1)3 + 5√2 (1)² – 7√2 = 2√2 + 5√2 – 7√2 = 7√2 – 7√2 = 0

Hence (x – 1) is a factor of p(x).

Ex. Without actual division, prove that the polynomial 2x³ + 13x² + x – 70 is exactly divisible by x – 2.

Sol. The polynomial p(x) = 2x³ + 13x² + x – 70 is exactly divisible by x – 2 means that x – 2 is a factor of

p(x) = 2x³ + 13x² + x – 70.

Now p(2) =2(2)³ + 13(2)² + 2 – 70 = 16 + 52 + 2 – 70 = 0

∴ By factor theorem, x – 2 is a factor of p(x)

i.e. p(x) = 2x³ + 13x² + x – 70 is exactly divisible by x – 2.

FACTORISING OF POLYNOMIAL OF HIGHER DEGREE

1. Obtain the polynomial p(x)
2. Obtain the constant term in p(x) and find its all possible factors. For example, in the polynomial x⁴ + x³ – 7x² – x + 6 the constant term is 6 and its factors are ±1, ±2, ±3, ±6.
3. 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.
4. Obtain the factors equal in number to the degree of polynomial. Let these are (x – a), (x – b), (x – c). ....
5. Write p(x) = k (x – a) (x – b) (x – c) ..... where k is constant.
6. Substitute any value of x other than a, b, c .......... and find the value of k.

Ex. Using factor theorem, factorize the polynomial x³ – 6x² + 11x – 6.
Sol. Let f(x) = x³ – 6x² + 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,x³ – 6x² + 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, x³ – 6x² + 11x – 6 = (x – 1) (x – 2) (x – 3)

Ex. Using factor theorem, factorize the polynomial x⁴ + x³ – 7x² – x + 6.

Sol. Let f(x) = x⁴ + x³ – 7x² – x + 6 the factors of constant term 6 are ±1, ±2, ±3 and ±6

Now, f(1) = 1 + 1 – 7 – 1 + 6 = 8 – 8 = 0

⇒ (x – 1) is a factor of f(x)

f(–1) = 1 – 1 – 7 + 1 + 6 = 8 – 8 = 0     ⇒     (x + 1) is a facor of f(x)

f(2) = 2⁴ + 2³ – 7 × 2² – 2 + 6 = 16 + 8 – 28 – 2 + 6 = 0

⇒ x – 2 is a factor of f(x)

f(–2) = (–2)⁴ + (–2)³ – 7(–2)² – (–2) + 6 = 16 – 8 – 28 + 2 + 6 = – 12    ≠ 0

⇒ x + 2 is not a factor of f(x)

f(–3) = (–3)⁴ + (–3)³ – 7(–3)² – (–3) + 6 = 81 – 27 – 63 + 3 + 6 = 90 – 90 = 0

⇒ x + 3 is a factor of f(x).

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

Thus, the factor 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)

⇒ x⁴ + x³ – 7x² – x + 6 = k (x – 1) (x + 1) (x – 2) (x + 3) ... (i)

Putting k = 0 on both sides, we get

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

Substituting k = 1 in (i), we get

x⁴ + x³ – 7x² – x + 6 = (x – 1) (x + 1) (x – 2) (x + 3)