Question

Factorise the following polynomials by splitting middle term:

$\sf \bold{1.} \: \: 6 {x}^{2} - x - 2 \\ \sf \bold{2. }\: \: 6 {p}^{2} + 17p + 12 \\ \sf \bold{3. }\: \: {y}^{2} - 4y - 21 \\ \sf \bold{4 . }\: \: {x}^{2} + 2 \sqrt{3} x - 24$
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Answer 1

Answer:-

1) 6x² - x - 2

In order to factorise a quadratic polynomial, we have to split the middle term through multiplying the first term and the constant at the last.

So, Product = 6 ( - 2) = - 12.

Now, We have to find the factors of - 12 whose sum is - 1 which is the coefficient of the x in middle term.

The factors of -12 whose sum is - 1 are (- 4) and ( + 3).

Hence; the polynomial can be written as:

⟹ 6x² - 4x + 3x - 2

Now, taking common factor of first terms and last two terms , we get;

⟹ 2x (3x - 2) + 1(3x - 2)

Now, taking 3x - 2 common we get,

⟹ (3x - 2)(2x + 1)

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2) 6p² + 17p + 12

Similarly;

Factors of (12)(6) = 72 whose sum is 17 are + 8 and + 9.

So,

⟹ 6p² + 8p + 9p + 12

⟹ 2p(3p + 4) + 3(3p + 4)

⟹ (3p + 4)(2p + 3)

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3) y² - 4y - 21

Factors of - 21 whose sum is - 4 are + 3 and - 7.

⟹ y² - 7y + 3y - 21

⟹ y(y - 7) + 3(y - 7)

⟹ (y - 7)(y + 3)

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4) x² + 2√3x - 24

Factors of - 24 whose sum is 2√3 are + 4√3 and - 2√3

[ ∵ (4√3)( - 2√3) = (4)(-2)(√3)(√3) = (- 8)(3) = - 24 ]

So,

⟹ x² + 4√3x - 2√3x - 24

⟹ x (x + 4√3) - 2√3( x + 4√3)

⟹ (x + 4√3)(x - 2√3)

Answer 2

Factorizing the following polynomials by middle term split :-

1. 6x² - x - 2

= 6x² - (4 - 3)x - 2

= 6x² - 4x + 3x - 2

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

= (3x - 2) (2x + 1) Ans.

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2. 6p² + 17p + 12

= 6p² + (9 + 8)p + 12

= 6p² + 9p + 8p + 12

= 3p(2p + 3) + 4(2p + 3)

= (2p + 3) (3p + 4) Ans.

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3. y² - 4y -21

= y² - (7 - 3)y -21

= y² - 7y + 3y -21

= y(y - 7) + 3(y - 7)

= (y - 7) (y + 3) Ans.

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4. x² + 2√3x - 24

= x² + 2√3x - 24

= x² + (4√3 - 2√3)x - 24

= x² + 4√3x - 2√3x - 24

= x(x + 4√3) - 2√3(x - 4√3)

= (x + 4√3) (x - 2√3) Ans.

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