Question
Factorise: $4 x^4+7 x^2-2$
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Answer
Let $x^2 = y$ Then, $4x^4 + 7x^2 - 2$
$= 4y^2 + 7y - 2$
$= 4y^2 + 8y - y - 2$
$= 4y(y + 2) - 1(y + 2)$
$= (y + 2)(4y - 1)$
Now replacing $y$ by $x^2,$
we get $4x^4 + 7x^2 - 2 = (x^2 + 2)(4x^2 - 1)$
Since $a^2 - b^2 = (a - b)(a + b) = (x^2 + 2)(2x + 1)(2x - 1)$
$= 4y^2 + 7y - 2$
$= 4y^2 + 8y - y - 2$
$= 4y(y + 2) - 1(y + 2)$
$= (y + 2)(4y - 1)$
Now replacing $y$ by $x^2,$
we get $4x^4 + 7x^2 - 2 = (x^2 + 2)(4x^2 - 1)$
Since $a^2 - b^2 = (a - b)(a + b) = (x^2 + 2)(2x + 1)(2x - 1)$
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