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Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b)²

px=x4+2x3-2x2...

Question

$p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ and $q (x) = x^{2} + 2 x - 3$ .
Then $p (x)$ is divisible by $q (x)$ , if we

Add

(x - 2)

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(x - 3)

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Add

(2 - x)

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(3 - x)

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Solution

The correct option is A Add $(x - 2)$
Given : $p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ , $q (x) = x^{2} + 2 x - 3$
To be exactly divisible $p (x)$ should be subtracted by $(- x + 2)$
i.e $p (x) = (x^{4} + 2 x^{3} - 2 x^{2} + x - 1) - (- x + 2)$
or,
$p (x) = (x^{4} + 2 x^{3} - 2 x^{2} + x - 1) + (x - 2)$

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v	?	$x^{2} + 2 x + 1$	$x^{4} - 2 x^{2} + 5 x - 7$	$4 x + 12$

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p(x)=x4+2x3−2x2+x−1 and q(x)=x2+2x−3.Then p(x) is divisible by q(x), if we

The correct option is A Add (x−2)Given : p(x)=x4+2x3−2x2+x−1,q(x)=x2+2x−3To be exactly divisible p(x) should be subtracted by (−x+2)i.e p(x)=(x4+2x3−2x2+x−1)−(−x+2)or, p(x)=(x4+2x3−2x2+x−1)+(x−2)

$p (x) = x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ and $q (x) = x^{2} + 2 x - 3$ .
Then $p (x)$ is divisible by $q (x)$ , if we