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.


Ver imagen InesWalston

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.