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Factorisation by Regrouping Terms

x3+6x+8=? a x...

Question

Factorize: $x^{2} + 6 x + 8 = ? (a) (x + 3) (x + 5) (b) (x + 3) (x + 4) (c) (x + 2) (x + 4) (d) (x + 1) (x + 8)$

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Solution

Given:
$x^{2} + 6 x + 8$
We need to select two terms such that their sum is ' $6 x$ ' and Their product is ' $8 x^{2}$ '.
Since, $4 x + 2 x = 6 x$ and $4 x \times 2 x = 8 x^{2}$
The two terms are $2 x$ and $4 x$ .
$\Rightarrow x^{2} + 6 x + 8$
$= x^{2} + 4 x + 2 x + 8$
$= x (x + 4) + 2 (x + 4)$
$= (x + 4) (x + 2)$
$∴ x^{2} + 6 x + 8 = (x + 4) (x + 2)$
Hence, Option C is correct.

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Similar questions

Q. Tick (✓) the correct answer:
x² + 6x + 8 = ?
(a) (x + 3)(x + 5)
(b) (x + 3)(x + 4)
(c) (x + 2)(x + 4)
(d) (x + 1)(x + 8)

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

Q. The factors of x³ − 7x + 6 are

(a) x (x − 6) (x − 1)

(b) (x² − 6) (x − 1)

(c) (x + 1) (x + 2) (x + 3)

(d) (x − 1) (x + 3) (x − 2)

Factorize: $x^{2} - 7 x + 12$

Factorise by splitting the middle term 2x² – 11x + 12

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Factorisation by Regrouping Terms

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