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factorise x4 +x2 +1

Let x2= p

Then x4= p2

x4+ x2+ 1

= p2+ p + 1

= p2+ 1 + p

= (p)2+ (1)2+ 2*p*1 - p

= (p)2+ (1)2+ 2p - p

= (p + 1)2- p

= (p + 1)2- (√p)2

= (p + 1 + √p) (p + 1 - √p)

Substituting the value of p,

( x2+ 1 + √x2) (x2+ 1 - √x2)

= (x2+ 1 +x) (x2+ 1 - x)

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View Full Answer
(X^2)^2+x^2+1
(x^2+x)^2+1
(x^2+x+1)^2
(x^2+x+1)(x^2+x+1)
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x4+x2+1
=(x2)2+2*x2*1+12
Now this is in the form of (a+b)2.
so, by factorising it we'll get, (x2+1)(x2+1)
Hope this helps you.
Regards.
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Ever tried Cymath? No, please do. You'll get your answer with steps. You're Welcome :)
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Please tell me define of Hardy number with examples.
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Historical Development of Cell Theory
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Consider x4 + x2 + 1
            = (x4 + 2x2 + 1) – x2
            = [(x2)2 + 2x2 + 1] – x2
            = [x2 + 1]2 – x2
It is in the form of (a2 – b2) = (a + b)(a – b)
Hence [x2 + 1]2 – x2 =  [x2 + 1 + x] [x2 + 1 – x]
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thank you  rohan 
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