Can you guys help me solve this 32 + 14x^2 -x^4
are you supposed to factor, or set equal to zero and solve?
factor
ok first we can try to factor \(32 + 14x -x^2\) and then replace \(x\) by \(x^2\)
there are a lot of factors of 32 but you know it has to look like \((a-x)(b+x)\) because that is the only way you are going to get an \(x^2\) out of it. so \(ab=32\) and \(a-b=14\) really you have to guess and check, but after some trial and error you should see that \(a=16\) and \(b=2\) will work because \(16\times 2=32\) and \(16-2=14\)
for x i've got x=4, x=-16
actually 2 :) my hand writing suck
ok good, first step would be to write as \[32+14x-x^2=(16-x)(2+x)\]
now we really have \((16-x^2)(2+x^2)\) so we can continue to factor, because \(16-x^2\) is the difference of two squares
final answer would then be \[(4+x)(4-x)(2+x^2)\]
tks
yw
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