I came across an expression, $8x^3-4x+1$. It was further factorized as $(2x-1)(4x^2+2x-1)$,i.e. They added and subtracted $4x^2$, split $-4x$ into $-2x&-2x$, took requisite common factors. But how do they think to proceed in this way, I cannot understand? I'd be glad if you help!
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1Divide the whole polynomial by 8(coefficient of $x^2$) and find all the factors of the constant term you get. One of them will be the zero of the polynomial. x=1/2 or x-1/2=0 or 2x-1 will be the factor. From there you have to do some rearrangements to remove the unnecessary terms you get. – piepi May 19 '16 at 12:36
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One simple trick is that a factor $ax+b$ must have $a$ divide the coefficient of the highest power and $b$ divide the constant term. So in this case you check to see if $\pm1,\pm\frac{1}{2},\pm\frac{1}{4}$ are roots. Of course, as soon as you find one, you factorise. – almagest May 19 '16 at 12:38
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The leading term is $8x^3$ so assuming integer coefficients, the leading terms of the two factors must be $(8x)(x^2)$ or $(x)(8x^2$ or $(2x)(4x^2)$ or $(4x^2)(2x)$. They tried each of those, saw they must be $(2x)(4x^2)$ and developed the rest from there. – user247327 May 19 '16 at 12:40