 # How can you determine all of the zeroes (real and 2022-01-17 Answered
How can you determine all of the zeroes (real and imaginary) of the polynomial function:
$P\left(x\right)={x}^{4}+2{x}^{3}-9{x}^{2}-10x-24?$
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Step 1
You can expect any factors to divide -24, the constant term over the lead coefficient. So check 1, 2, 3, 4, 6,12, 24 and their negatives. Turns out that 3 and -4 work, so you are down to a quadratic
$P\left(x\right)=\left(x-3\right)×\left(x+4\right)×\left({x}^{2}+x+2\right)$
so the roots are

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Step 1
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
${x}^{4}+2{x}^{3}-9{x}^{2}-10x-24=\left({x}^{2}+x+k{\right)}^{2}-\left(mx+n{\right)}^{2}$
that is:
${x}^{4}+2{x}^{3}+\left(1+2k-{m}^{2}\right){x}^{2}+2\left(k-mn\right)x+\left({k}^{2}-{n}^{2}\right)={x}^{4}+2{x}^{3}-9{x}^{2}-10x-24$
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
$1+2k-{m}^{2}=-9\phantom{\rule{0ex}{0ex}}2k-2mn=-10\phantom{\rule{0ex}{0ex}}{k}^{2}-{n}^{2}=-24$
Rearrange these equations and get
${m}^{2}=2\left(k+5\right)\phantom{\rule{0ex}{0ex}}mn=k+5\phantom{\rule{0ex}{0ex}}{n}^{2}={k}^{2}+24$
From the 2 equations on the left we get
${m}^{2}=2mn$
and therefore one possible solution would be:
$m=0\phantom{\rule{0ex}{0ex}}k=-5\phantom{\rule{0ex}{0ex}}{n}^{2}=49={7}^{2}$
Hence we get the following factorization of the given polynomial:
${x}^{4}+2{x}^{3}-9{x}^{2}-10x-24=\left({x}^{2}+x-5{\right)}^{2}-{7}^{2}=\phantom{\rule{0ex}{0ex}}=\left({x}^{2}+x+2\right)\left({x}^{2}+x-12\right)$
So, all the problem has now been reduced into finding the 4 roots of the 2 quadratic polynomials, which are
${x}_{1}=-\frac{1}{2}+\frac{\sqrt{7}}{2}i\phantom{\rule{0ex}{0ex}}{x}_{2}=-\frac{1}{2}-\frac{\sqrt{7}}{2}i$
and
${x}_{3}=-\frac{1}{2}+\frac{7}{2}=3\phantom{\rule{0ex}{0ex}}{x}_{4}=-\frac{1}{2}-\frac{7}{2}=-4$