Question

1. given $f(x)=x^4 + 6x^3 + 4x^2 - 24x - 32$, where $(x^2 + 6x + 8)$ is a factor of the polynomial, find all zeros and write the polynomial in fully factored form.
2. write a polynomial in simplest standard form with real coefficients whose zeros are $sqrt{3}$ and 7.

Explanation:

Problem 30:

Step 1: Factor the given quadratic factor

First, factor \(x^{2}+6x + 8\). We need two numbers that multiply to \(8\) and add to \(6\). Those numbers are \(2\) and \(4\), so \(x^{2}+6x + 8=(x + 2)(x+4)\).

Step 2: Perform polynomial division or use synthetic division to divide \(f(x)=x^{4}+6x^{3}+4x^{2}-24x - 32\) by \(x^{2}+6x + 8\)

Let's use polynomial long division. Divide \(x^{4}+6x^{3}+4x^{2}-24x - 32\) by \(x^{2}+6x + 8\).

\(x^{4}\div x^{2}=x^{2}\), multiply \(x^{2}+6x + 8\) by \(x^{2}\) to get \(x^{4}+6x^{3}+8x^{2}\). Subtract this from \(x^{4}+6x^{3}+4x^{2}-24x - 32\):

\((x^{4}+6x^{3}+4x^{2}-24x - 32)-(x^{4}+6x^{3}+8x^{2})=- 4x^{2}-24x - 32\)

Now, divide \(-4x^{2}\) by \(x^{2}\) to get \(-4\). Multiply \(x^{2}+6x + 8\) by \(-4\) to get \(-4x^{2}-24x - 32\). Subtract this from \(-4x^{2}-24x - 32\):

\((-4x^{2}-24x - 32)-(-4x^{2}-24x - 32)=0\)

So, \(f(x)=(x^{2}+6x + 8)(x^{2}-4)\)

Step 3: Factor the remaining quadratic

Factor \(x^{2}-4\) using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a=x\) and \(b = 2\). So \(x^{2}-4=(x + 2)(x - 2)\)

Step 4: Combine all factors and find zeros

We already factored \(x^{2}+6x + 8=(x + 2)(x + 4)\), so \(f(x)=(x + 2)(x + 4)(x + 2)(x - 2)=(x + 2)^{2}(x + 4)(x - 2)\)

To find the zeros, set \(f(x)=0\):

\((x + 2)^{2}(x + 4)(x - 2)=0\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So:

\(x+2 = 0\) gives \(x=-2\) (with multiplicity \(2\))

\(x + 4=0\) gives \(x=-4\)

\(x - 2=0\) gives \(x = 2\)

Problem 31:

Step 1: Recall the conjugate root theorem

For a polynomial with real coefficients, if \(\sqrt{3}\) is a zero, then its conjugate \(-\sqrt{3}\) must also be a zero (since the coefficients are real and \(\sqrt{3}\) is irrational).

Step 2: Write the polynomial in factored form

If the zeros are \(x=\sqrt{3}\), \(x =-\sqrt{3}\) and \(x = 7\), then the polynomial in factored form is \(f(x)=(x-\sqrt{3})(x+\sqrt{3})(x - 7)\)

Step 3: Multiply the factors

First, multiply \((x-\sqrt{3})(x+\sqrt{3})\) using the difference of squares formula \((a - b)(a + b)=a^{2}-b^{2}\), where \(a=x\) and \(b=\sqrt{3}\). So \((x-\sqrt{3})(x+\sqrt{3})=x^{2}-3\)

Then, multiply \((x^{2}-3)(x - 7)\):

$$
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Answer:

s:

Problem 30:

• Zeros: \(x=-4\), \(x=-2\) (multiplicity \(2\)), \(x = 2\)
• Fully factored form: \(f(x)=(x + 4)(x + 2)^{2}(x - 2)\)

Problem 31:

The polynomial in standard form is \(x^{3}-7x^{2}-3x + 21\)