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