Respuesta :
Answer-
Zeros of the functions are 0 with multiplicity of 2 and 1, 3 with multiplicity of 1.
Solution-
The given polynomial is,
[tex]x^4-4x^3+3x^2[/tex]
For calculating roots,
[tex]\Rightarrow x^4-4x^3+3x^2=0[/tex]
[tex]\Rightarrow x^4-3x^3-x^3+3x^2=0[/tex]
[tex]\Rightarrow x(x^3-3x^2)-1(x^3-3x^2)=0[/tex]
[tex]\Rightarrow (x-1)(x^3-3x^2)=0[/tex]
[tex]\Rightarrow (x-1)(x^2(x-3))=0[/tex]
[tex]\Rightarrow (x^2)(x-1)(x-3)=0[/tex]
The root at x²=0, has multiplicity of 2 as the power is 2.
So, the graph will bounce at x=0
[tex]\Rightarrow x^2=0,x-1=0,x-3=0[/tex]
[tex]\Rightarrow x=0,1,3[/tex]
At x=1, x=3 the graph will go through the x axis. They have multiplicity of 1.

The zeroes of the function are [tex]\boxed{0, 1 \text{and } 3}[/tex]. Multiplicity of [tex]0[/tex] is [tex]\boxed{\bf 2}[/tex], multiplicity of [tex]1[/tex] is [tex]\boxed{\bf 1}[/tex] and the multiplicity of [tex]3[/tex] is [tex]\boxed{\bf 1}[/tex].
Further explanation:
The given polynomial is [tex]x^{4}-4x^{3}+3x^{2}[/tex].
The above polynomial is a biquadratic polynomial.
Consider the given polynomial as follows:
[tex]\boxed{P(x)=x^{4}-4x^{3}+3x^{2}}[/tex]
The polynomial [tex]P(x)[/tex] is a polynomial of degree [tex]4[/tex].
To find the zeroes of polynomial [tex]P(x)[/tex] equate this polynomial to 0.
[tex]\begin{aligned}P(x)&=0\\x^{4}-4x^{3}+3x^{2}&=0\end{aligned}[/tex]
Factorize the polynomial [tex]P(x)[/tex] by taking common term [tex]x^{2}[/tex], since [tex]x^{2}[/tex] is the only lowest term in above polynomial with power [tex]2[/tex].
Factorize the polynomial [tex]P(x)[/tex] as follows:
[tex]x^{2}(x^{2}-4x+3)=0[/tex]
Now, factorize the quadratic term [tex]x^{2}-4x+3[/tex] as follows:
[tex]\begin{aligned}x^{2}-4x+3&=x^{2}-3x-x+3\\&=x(x-3)-1(x-3)\\&=(x-1)(x-3)\end{aligned}[/tex]
Substitute [tex](x-1)(x-3)[/tex] in the polynomial [tex]P(x)[/tex] as shown below:
[tex]x^{2}(x-1)(x-3)=0[/tex]
Now, from zero-product property the zeroes of the polynomial are shown below:
[tex]\begin{aligned}x^{2}(x-1)(x-3)&=0\\x&=0,1,3\end{aligned}[/tex]
Here, [tex]0[/tex] is a zero of multiplicity [tex]2[/tex], since the power of [tex]x[/tex] is [tex]2[/tex].
Since, the degree of the polynomial is [tex]4[/tex], therefore, there will be [tex]4[/tex] solution to this polynomial.
Thus, the zeroes of the function [tex]P(x)=x^{4}-4x^{3}+3x^{2}[/tex] are [tex]\bf 0, 1 \text{and } 3[/tex] with [tex]0[/tex] having multiplicity of [tex]\bf 2[/tex] and both [tex]1[/tex] and [tex]3[/tex] have multiplicity of [tex]\bf 1[/tex].
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Answer details:
Grade: High school
Subject: Mathematics
Chapter: Polynomial
Keywords: Zeroes, function, multiplicity, x^4-4x^3+3x^2, degree, highest power, polynomial, quadratic, equation, zero-product property, factorization, biquadratic expression.