Question

which of the following describes the zeroes of the graph of $f(x) = 3x^6 + 30x^5 + 75x^4$?

• $-5$ with multiplicity $2$ and $\frac{1}{3}$ with multiplicity $4$
• $5$ with multiplicity $2$ and $\frac{1}{3}$ with multiplicity $4$
• $-5$ with multiplicity $2$ and $0$ with multiplicity $4$
• $5$ with multiplicity $2$ and $0$ with multiplicity $4$

Explanation:

Step1: Factor the polynomial

First, factor out the greatest common factor (GCF) from \( f(x) = 3x^6 + 30x^5 + 75x^4 \). The GCF of \( 3x^6 \), \( 30x^5 \), and \( 75x^4 \) is \( 3x^4 \). So we have:
\( f(x)=3x^4(x^2 + 10x + 25) \)

Step2: Factor the quadratic

Now, factor the quadratic \( x^2 + 10x + 25 \). Notice that it is a perfect square trinomial, since \( 10x = 2\times5\times x \) and \( 25 = 5^2 \). So \( x^2 + 10x + 25=(x + 5)^2 \)

Step3: Write the fully factored form

Substituting back, we get \( f(x)=3x^4(x + 5)^2 \)

Step4: Find the zeroes and their multiplicities

To find the zeroes, set \( f(x) = 0 \). Then \( 3x^4(x + 5)^2 = 0 \). Using the zero - product property, we have \( x^4 = 0 \) or \( (x + 5)^2 = 0 \)

• For \( x^4 = 0 \), the solution is \( x = 0 \), and the multiplicity is 4 (because the exponent of \( x \) in \( x^4 \) is 4).
• For \( (x + 5)^2 = 0 \), we solve \( x+5 = 0 \), so \( x=-5 \), and the multiplicity is 2 (because the exponent of \( (x + 5) \) is 2).

Answer:

\(-5\) with multiplicity \(2\) and \(0\) with multiplicity \(4\) (corresponding to the third option: \(-5\) with multiplicity \(2\) and \(0\) with multiplicity \(4\))