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**FACTOR THEOREM**

If p(x) is any polynomial of degree 1 or greater than 1 and ‘a’ be any real number such that: (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).**FACTORIZATION OF POLYNOMIALS:**

[Using the “splitting the middle-term” method] Let ax^{2} + bx + c be a quadratic polynomial such that a, b and c are constants and a ≠ 0.

To factorise it, we split the co-efficient of x into two parts ‘l’ and ‘m’ such that l + m = b

l x m = a x c**Ques 1: Determine which of the following polynomials has a factor (x + 1): ****(i) x ^{3} + x^{2} + x + 1 **

∴ The zero of x + 1 is –1.

(i) p(x) = x^{3} + x^{2} + x + 1

∴ p(–1) = (–1)^{3 }+ (–1)^{2} + (–1) + 1

= –1 + 1 – 1 + 1 = 0

i.e. when p(x) is divided by (x + 1), then the remainder is zero.

∴ (x + 1) is a factor of x^{3 }+ x^{2} + x + 1.

(ii) ∵ p(x) = x^{4} + x^{3} + x^{2} + x + 1

∴ p(–1) = (–1)^{4} + (–1)^{3} + (–1)^{2} + (–1) + 1

= 1 – 1 + 1 – 1 + 1

= 3 – 2 = 1

∵ f(–1) ≠ 0

∴ p(x) is not divisible by x + 1. i.e. (x + 1) is not a factor of x^{4} + x^{3} + x^{2} + x + 1.

(iii) ∵ p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

∴ p(–1) = (–1)^{4} + 3(–1)^{3} + 3(–1)^{2} + (–1) + 1 = (1) + 3(–1) + 3(1) + (–1) + 1

= 1 – 3 + 3 – 1 + 1

= 1 ≠ 0

∵ f(–1) ≠ 0

∴ (x + 1) is not a factor of x^{4} + 3x^{3} + 3x^{2 }+ x + 1.

(iv) ∵ p(x) = x^{3} – x^{2} – (2 + √2) x + √2)

∴ p(–1) = (–1)^{3} – (–1)^{2} – (2 + 2) (–1) + 2 = –1 – 1 – (–1) (2 + √2) + 2

= –1 – 1 + 1 (2 + √2) + 2

= –1 – 1 + 2 + √2 + √2)

= –2 + 2 + 2√2

= 2√2 ≠ 0

Since p(–1) ≠ 0.

∴ (x + 1) is not a factor of x^{4} + 3x^{3} + 3x^{2 }+ x + 1.**Ques 2: Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases: ****(i) p(x) = 2x ^{3} + x^{2} – 2x – 1, g(x) = x + 1 **

**Ans: **(i) We have p(x) = 2x^{3} + x^{2} – 2x – 1 and g(x) = x + 1

∴ p(–1) = 2(–1)^{3 }+ (–1)^{2} – 2(–1) –1

= 2(–1) + 1 + 2 – 1

= –2 + 1 + 2 – 1

= –3 + 3 = 0

∵ p(–1) = 0

∴ g(x) is a factor of p(x).

(ii) We have p(x) = x^{3} + 3x^{2} + 3x + 1 and g(x) = x + 2

∴ p(–2) = (–2)^{3} + 3(–2)^{2} + 3(–2) + 1 = –8 + 3(4) + (–6) + 1

= –8 + 12 – 6 + 1

= –8 – 6 + 12 + 1

= –14 + 13 = –1

∴ p(–2) ≠ 0

Thus, g(x) is not a factor of p(x).

(iii) We have p(x) = x^{3 }– 4x^{2} + x + 6 and g(x) = x – 3

∴ p(3) = (3)^{3} – 4 (3)^{2} + (3) + 6 = 27 –4(9) + 3 + 6

= 27 – 36 + 3 + 6 = 0

Since g(x) = 0

∴ g(x) is a factor of p(x).**Ques 3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: ****(i) p(x) = x ^{2} + x + k **

For x – 1 be a factor of p(x), p(1) should be equal to 0.

We have p(1) = (1)

or p(1) = 1 + 1 + k = k + 2

∴ k + 2 = 0 ⇒ k= –2

(ii) Here, p(x) = 2x^{2} + kx + √2

For x – 1 be a factor of p(x), p(1) = 0

Since, p(1) = 2(1)^{2 }+ k(1) + √2

= 2 + k + √2

∵ p(1) must be equal to 0.

∴ k + 2 + √2 = 0

⇒ k = –2 – √2

or k = – (2 + √2) .

(iii) Here p(x) = kx^{2} – √2 x + 1 and g(x) = x – 1

∴ For (x – 1) be a factor of p(x), p(1) should be equal to 0.

Since p(1) = k(1)^{2} – √2 (1) + 1

or p(1) = k – √2 + 1

or p(1) = k – √2 + 1

∴ k – √2 +1 = 0 ⇒ k= √2 – 1

(iv) Here p(x) = kx^{2} – 3x + k and g(x) = x – 1

For g(x) be a factor of p(x), p(1) should be equal to 0.

Since p(1) = k(1)^{2} – 3(1) + k = k – 3 + k = 2k – 3

∴ 2k – 3 = 0 ⇒ k = 3/2**Ques 4: Factorise: ****(i) 12x ^{2 }– 7x + 1 **

Here co-efficient of x

Co-efficient of x = –7 and constant term = 1

∴ a = 12, b = –7, c = 1

Now, l + m = –7 and lm = ac = 12 x 1

∴ We have l = (–4) and m = (–3) i.e. b = –7 = (–4 – 3).

Now, 12x

= 4x(3x – 1) – 1(3x – 1)

= (3x – 1)(4x – 1)

Thus, 12x

(ii) 2x^{2} + 7x + 3 Here, a = 2, b = 7 and c = 3

∴ l + m = 7 and lm = 2 x 3 = 6

i.e. 1 + 6 = 7 and 1 x 6 = 6

∴ l = 1 and m = 6

We have 2x^{2} + 7x + 3 = 2x^{2} + x + 6x + 3

= x(2x + 1) + 3(2x + 1)

= (2x + 1)(x + 3)

Thus, 2x^{2} + 7x + 3 = (2x + 1)(x + 3)

(iii) 6x^{2} + 5x – 6

We have a = 6, b = 5 and c = –6

∴ l + m = 5 and lm = ac = 6 x (–6) = –36

∴ l + m = 9 + (–4)

∴ 6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3)(3x – 2)

Thus, 6x^{2} + 5x – 6 = (2x + 3)(3x – 2)

(iv)3x^{2} – x – 4

We have a = 3, b = –1 and c = –4

∴ l + m = –1 and lm = 3 x (–4) = –12

∴ l = –4 and m = 3

Now, 3x^{2} – x – 4 = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4)(x + 1)

Thus, 3x^{2} – x – 4 = (3x – 4)(x + 1)**Ques 5: Factorise: ****(i) x ^{3 }– 2x^{2} – x + 2 **

Rearranging the terms, we have x^{3} – 2x^{2} – x + 2 = x^{3} – x – 2x^{2} + 2

= x(x^{2} – 1) – 2(x^{2} – 1)

= (x^{2} – 1)(x – 2)

= [(x)^{2} – (1)^{2}]

[x – 2] = (x – 1)(x + 1)(x – 2) [∵ a^{2} – b^{2} = (a + b)(a – b)]

Thus, x^{3} – 2x^{2} – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) x^{3 }– 3x^{2} – 9x – 5

We have p(x) = x^{3} – 3x^{2} – 9x – 5

By trial, let us find: p(1) = (1)^{3} – 3(1)^{2} – 9(1) – 5 = 3 – 3 – 9 – 5

= –14 ≠ 0

Now p(–1) = (–1)3 – 3(–1)^{2} – 9(–1) –5 =

–1 – 3(1) + 9 – 5

= –1 – 3 + 9 – 5 = 0

∴ By factor theorem, [x – (–1)] is a factor of p(x).

∴ x^{2 }– 3x^{2} – 9x – 5 = (x + 1)(x^{2} – 4x – 5)

= (x + 1)[x^{2} – 5x + x – 5] [Splitting –4 into –5 and +1]

= (x + 1) [x(x – 5) + 1(x – 5)]

= (x + 1) [(x – 5) (x + 1)]

= (x + 1)(x – 5)(x + 1)

(iii) x^{3} + 13x^{2} + 32x + 20

We have p(x) = x^{3} + 13x^{2} + 32x + 20

By trial, let us find: p(1) = (1)^{3 }+ 13(1)^{2 }+ 32(1) + 20

= 1 + 13 + 32 + 20

= 66 ≠ 0

Now p(–1) = (–1)^{3} + 13(–1)^{2 }+ 32(–1) + 20

= –1 + 13 – 32 + 20

= 0

∴ By factor theorem, [x – (–1)], i.e. (x + 1) is a factor p(x).

or x^{3} + 13x^{2 }+ 32x + 20

= (x + 1)(x^{2} + 12x + 20)

= (x + 1)[x^{2 }+ 2x + 10x + 20] [Splitting the middle term]

= (x + 1)[x(x + 2) + 10(x + 2)]

= (x + 1)[(x + 2)(x + 10)]

= (x + 1)(x + 2)(x + 10)

(iv) 2y^{3} + y^{2} – 2y – 1

We have p(y) = 2y^{3} + y^{2} – 2y – 1

By trial, we have p(1) = 2(1)^{3} + (1)^{2} – 2(1) – 1

= 2(1) + 1 – 2 – 1

= 2 + 1 – 2 – 1 = 0

∴ By factor theorem, (y – 1) is a factor of p(y).

∴ 2y^{3} – y^{2 }– 2y – 1 = (y – 1)(2y^{2} + 3y + 1)

= (y – 1)[2y^{2} + 2y + y + 1] [Splitting the middle term]

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)[(y + 1)(2y + 1)]

= (y – 1)(y + 1)(2y + 1)

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