The correct option is B (2x^2+x+1)(2x^2 x+1) Given, 4x^4+3x^2+1 ⇒ 4x^4+4x^2 x^2+1⇒ #160; 4x^4+4x^2+1 x^2 ∵ (a+b #160; )^2=a^2+2ab+b^2 ...

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Factorisation of Quadratic Polynomials - Factor Theorem

Factorise: ...

Question

Factorise: $4 x^{4} + 3 x^{2} + 1$

(2 x^{2} + x + 1) (2 x^{2} + x - 1)

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(2 x^{2} + x + 1) (2 x^{2} - x + 1)

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(2 x^{2} + x - 1) (2 x^{2} - x - 1)

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(2 x^{2} - x + 1) (2 x^{2} - x - 1)

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Solution

The correct option is B $(2 x^{2} + x + 1) (2 x^{2} - x + 1)$
Given, $4 x^{4} + 3 x^{2} + 1$
$\Rightarrow 4 x^{4} + 4 x^{2} - x^{2} + 1 \Rightarrow 4 x^{4} + 4 x^{2} + 1 - x^{2}$
$∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
$\Rightarrow {(2 x^{2} + 1)}^{2} - {(x)}^{2}$
$∵ a^{2} - b^{2} = (a + b) (a - b)$
$\Rightarrow (2 x^{2} + 1 + x) (2 x^{2} + 1 - x)$
$\Rightarrow (2 x^{2} + x + 1) (2 x^{2} - x + 1)$

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Similar questions

Factorise:

(i) 12x² − 7x + 1 (ii) 2x² + 7x + 3

(iii) 6x² + 5x − 6 (iv) 3x² − x − 4

\frac{3 x + 2}{(x + 1) (2 x^{2} + 3)} = \frac{A}{(x + 1)} + \frac{B x + C}{(2 x^{2} + 3)}

A + C - B =

12 x^{2} - 7 x + 1

2 x^{2} + 7 x + 3

6 x^{2} + 5 x - 6

3 x^{2} - x - 4

Factorise:
(i) $12 x^{2} - 7 x + 1$
(ii) $2 x^{2} + 7 x + 3$
(iii) $6 x^{2} + 5 x - 6$
(iv) $3 x^{2} - x - 4$

(2 x^{2} - 3 x + 1) (2 x^{2} + 5 x + 1) = 9 x^{2}

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