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Answer :
The zeros of the equation [tex]{x^5}-3{x^4}-15{x^3}+45{x^2}-16x+48=0[/tex] are [tex]\boxed{--4,{\text{ }}3,{\text{ }}4,+i{\text{ and}}-i}[/tex]
Further explanation:
The Fundamental Theorem of Algebra states that the polynomial has n roots if the degree of the polynomial is n.
[tex]f\left(x\right)=a{x^n}+b{x^{n-1}}+\ldots+cx+d[/tex]
The polynomial function has n roots or zeroes.
Given:
The polynomial function is [tex]{x^5}-3{x^4}-15{x^3}+45{x^2}-16x+48=0[/tex].
Explanation:
The polynomial function [tex]{x^5}-3{x^4}-15{x^3}+45{x^2}-16x+48=0[/tex] has five zeroes as the degree of the polynomial is 5.
Solve the equation to obtain the zeroes.
[tex]\begin{aligned}\left({{x^5}-3{x^4}}\right)+\left({-15{x^3}+45{x^2}}\right)+\left({-16x+48}\right)&=0\\{x^4}\left({x-3}\right)-15{x^2}\left({x-3}\right)-16\left({x-3}\right)&=0\\\left({x-3}\right)\left({{x^4}-15{x^2}-16}\right)&=0\\\end{aligned}[/tex]
The first zero of the equation can be calculated as follows,
[tex]\begin{aligned}x-3&=0\\x&=3\\\end{aligned}[/tex]
Now solve the equation [tex]{x^4}-15{x^2}[/tex]-16 to obtain the remaining zeros.
[tex]\begin{aligned}{x^4}-15{x^2}-16&=0\\\left({{x^2}-16}\right)\left({{x^2}+1}\right)&=0\\\end{aligned}[/tex]
Substitute [tex]{x^2}[/tex]-16 equal to zero to obtain the zeroes.
[tex]\begin{aligned}{x^2}-16&=0\\{x^2}&=16\\x&=\sqrt{16}\\x&=\pm4\\\end{aligned}[/tex]
There zeroes are 4 and -4.
Now substitute [tex]{x^2}[/tex]+1 equal to zero to obtain the zeros.
[tex]\begin{aligned}{x^2}+1&=0\\{x^2}&=-1\\x&=\sqrt{-1}\\x&=\pm i\\\end{aligned}[/tex]
Hence, the zeros of the equation [tex]{x^5}-3{x^4}-15{x^3}+45{x^2}-16x+48=0[/tex] are [tex]\boxed{--4,{\text{ }}3,{\text{ }}4,+i{\text{ and}}-i}[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Polynomials
Keywords: quadratic equation, equation factorization. Factorized form, polynomial, quadratic formula, zeroes, Fundamental Theorem of algebra, polynomial, six roots.
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Rewritten by : Barada
The zeros of the equation [tex]x^5-3x^4-15x^3+45x^2-16x+48=0[/tex] are [tex]\pm4,3,\pm\iota[/tex].
The given Equation is,
[tex]x^5-3x^4-15x^3+45x^2-16x+48=0[/tex]
The number of zeroes depends upon degree of equation as this equation has degree [tex]5[/tex] so number of zeroes are
Solve the equation:
[tex]x^5-3x^4-15x^3+45x^2-16x+48=0[/tex]
[tex]x^4(x-3)+(-15x^3+45x^2)+(-16x+48)=0[/tex]
[tex]x^4(x-3)-15x^2(x-3)-16(x-3)=0[/tex]
Take [tex](x-3)[/tex] common , we get
[tex](x-3)(x^4-15x^2-16)=0[/tex]
[tex]x-3=0\\x=3[/tex]
Also,
[tex]x^4-15x^2-16=0\\\\x^4-16x^2+x^2-16=0\\\\(x^2-16)(x^2+1)=0[/tex]
[tex]x^2-16=0\\x^2=16\\x=\pm4[/tex]
Also,
[tex]x^2+1=0\\x^2=-1\\x=\pm\iota[/tex]
Hence the equation has [tex]5[/tex] roots.
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