Question

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.

Factors and multiples
ANSWERED
asked 2021-08-07
Given \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}\), 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 \(\displaystyle{P}{\left({x}\right)}=\)

Expert Answers (1)

2021-08-08
Step 1
Given: \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}\)
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, \(\displaystyle{\left({x}-{4}{i}\right)},{\left({x}+{4}{i}\right)}\) are factors of P(x)
So, \(\displaystyle{\left({x}-{4}{i}\right)}{\left({x}+{4}{i}\right)}\) is a factor of P(x)
\(\displaystyle\Rightarrow{x}^{{{2}}}+{16}\) is a factor of P(x)
Dividing P(x) by \(\displaystyle{x}^{{{2}}}+{16}\)
So, \(\displaystyle{\frac{{{3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}}}{{{x}^{{{2}}}+{16}}}}={3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\)
\(\displaystyle{P}{\left({x}\right)}={\left({3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\right)}{\left({x}^{{{2}}}+{16}\right)}\)
Step 2
Consider \(\displaystyle{F}{\left({x}\right)}={3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\)
We observe that \(\displaystyle{F}{\left(-{2}\right)}={3}{\left(-{2}\right)}^{{{3}}}-{8}{\left(-{2}\right)}^{{{2}}}-{13}{\left(-{2}\right)}+{30}\)
\(\displaystyle{F}{\left(-{2}\right)}=-{24}-{32}+{26}+{30}\)
\(\displaystyle{F}{\left(-{2}\right)}={0}\)
Similarly, \(\displaystyle{F}{\left({3}\right)}={0}\)
So, -2,3 are zeros of F(x) and hence of P(x) too
Let \(\displaystyle\alpha\) be the fifth zero of P(x)
We know that the sum of zeros of \(\displaystyle{P}{\left({x}\right)}=\frac{{8}}{{3}}\)
\(\displaystyle\alpha+{4}{i}-{4}{i}+{\left(-{2}\right)}+{\left({3}\right)}={\frac{{{8}}}{{{3}}}}\)
\(\displaystyle\alpha={\frac{{{8}}}{{{3}}}}-{1}\)
\(\displaystyle\alpha={\frac{{{5}}}{{{3}}}}\)
Hence, the zeros are \(\displaystyle-{2},{3},{4}{i},-{4}{i},{\frac{{{5}}}{{{3}}}}\)
Hence, \(\displaystyle{P}{\left({x}\right)}={3}{\left({x}+{2}\right)}{\left({x}-{3}\right)}{\left({x}-{4}{i}\right)}{\left({x}+{4}{i}\right)}{\left({x}-{\frac{{{5}}}{{{3}}}}\right)}\)
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