# Given P(x)=3x^{5}-8x^{4}+35x^{3}-98x^{2}-208x+480, and that 4i is a zero, write P in factored form be sure to write the full equation.

Given $P\left(x\right)=3{x}^{5}-8{x}^{4}+35{x}^{3}-98{x}^{2}-208x+480$, and that 4i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including $P\left(x\right)=$
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Step 1
Given: $P\left(x\right)=3{x}^{5}-8{x}^{4}+35{x}^{3}-98{x}^{2}-208x+480$
It is given that 4i is a zero of P(x)
We know that for polynomials with real coefficients, the complex zeros always exists in conjugate pairs
Hence, -4i is also a zero of P(x)
Hence, $\left(x-4i\right),\left(x+4i\right)$ are factors of P(x)
So, $\left(x-4i\right)\left(x+4i\right)$ is a factor of P(x)
$⇒{x}^{2}+16$ is a factor of P(x)
Dividing P(x) by ${x}^{2}+16$
So, $\frac{3{x}^{5}-8{x}^{4}+35{x}^{3}-98{x}^{2}-208x+480}{{x}^{2}+16}=3{x}^{3}-8{x}^{2}-13x+30$
$P\left(x\right)=\left(3{x}^{3}-8{x}^{2}-13x+30\right)\left({x}^{2}+16\right)$
Step 2
Consider $F\left(x\right)=3{x}^{3}-8{x}^{2}-13x+30$
We observe that $F\left(-2\right)=3{\left(-2\right)}^{3}-8{\left(-2\right)}^{2}-13\left(-2\right)+30$
$F\left(-2\right)=-24-32+26+30$
$F\left(-2\right)=0$
Similarly, $F\left(3\right)=0$
So, -2,3 are zeros of F(x) and hence of P(x) too
Let $\alpha$ be the fifth zero of P(x)
We know that the sum of zeros of $P\left(x\right)=\frac{8}{3}$
$\alpha +4i-4i+\left(-2\right)+\left(3\right)=\frac{8}{3}$
$\alpha =\frac{8}{3}-1$
$\alpha =\frac{5}{3}$
Hence, the zeros are $-2,3,4i,-4i,\frac{5}{3}$
Hence, $P\left(x\right)=3\left(x+2\right)\left(x-3\right)\left(x-4i\right)\left(x+4i\right)\left(x-\frac{5}{3}\right)$