Quadratic and Polynomial Functions $\newline$ Using a given zero to write a polynomial as a product of line $\newline$ Try Again $\newline$ Your answer Is Incorrect. $\newline$ For the polynomial below, $3$ is a zero. $\newline$ $h(x)=x^{3}+3x^{2}-16x-6$ $\newline$ Express $\newline$ $h(x)$ as a product of linear factors. $\newline$ $h(x)=\boxed{\phantom{answer}}$

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Q. Quadratic and Polynomial Functions

\newline

Using a given zero to write a polynomial as a product of line

\newline

Try Again

\newline

Your answer Is Incorrect.

\newline

For the polynomial below,

3

is a zero.

\newline

h(x)=x^{3}+3x^{2}-16x-6

\newline

Express

\newline

h(x)

as a product of linear factors.

\newline

h(x)=\boxed{\phantom{answer}}

Confirm Zero of Polynomial: First, we need to confirm that $3$ is indeed a zero of the polynomial $h(x)$ by substituting $x$ with $3$ and checking if $h(3)$ equals zero. $\newline$ Calculation: $h(3) = 3^3 + 3*(3)^2 - 16*3 - 6 = 27 + 27 - 48 - 6 = 54 - 54 = 0$
Perform Polynomial Division: Since $3$ is a zero, $(x - 3)$ is a factor of $h(x)$ . We will perform polynomial division to divide $h(x)$ by $(x - 3)$ to find the other factors.
Set Up Long Division: Performing the division, we set up the long division of $h(x)$ by $(x - 3)$ .
Divide First Term: Divide the first term of $h(x)$ , which is $x^3$ , by the first term of $(x - 3)$ , which is $x$ , to get $x^2$ . Multiply $(x - 3)$ by $x^2$ and subtract the result from $h(x)$ . $\newline$ Calculation: $x^3$ divided by $x$ is $x^2$ , so we have $x^3$ $1$ . Subtracting this from $h(x)$ gives us $x^3$ $3$ , which simplifies to $x^3$ $4$ .
Repeat Division Process: Repeat the division process with the new polynomial $6x^2 - 16x - 6$ . Divide the first term, $6x^2$ , by $x$ to get $6x$ . Multiply $(x - 3)$ by $6x$ and subtract the result from $6x^2 - 16x - 6$ . Calculation: $6x^2$ divided by $x$ is $6x$ , so we have $6x^2$ $0$ . Subtracting this from $6x^2 - 16x - 6$ gives us $6x^2$ $2$ , which simplifies to $6x^2$ $3$ .
Divide Remaining Polynomial: Finally, divide the remaining polynomial $2x - 6$ by $x - 3$ . Divide the first term, $2x$ , by $x$ to get $2$ . Multiply $(x - 3)$ by $2$ and subtract the result from $2x - 6$ . $\newline$ Calculation: $2x$ divided by $x$ is $2$ , so we have $x - 3$ $1$ . Subtracting this from $2x - 6$ gives us $x - 3$ $3$ , which means the division is complete and there is no remainder.
Find Quadratic Factors: The result of the division gives us the other factors of $h(x)$ . Since we divided $h(x)$ by $(x - 3)$ and got $x^2 + 6x + 2$ as the quotient, $h(x)$ can be expressed as $(x - 3)(x^2 + 6x + 2)$ .
Use Quadratic Formula: Now we need to factor the quadratic $x^2 + 6x + 2$ . We look for two numbers that multiply to $2$ and add up to $6$ . Unfortunately, there are no two such real numbers, which means the quadratic does not factor over the real numbers. We can use the quadratic formula to find the zeros of the quadratic.
Substitute into Formula: The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = 6$ , and $c = 2$ for the quadratic $x^2 + 6x + 2$ .
Calculate Zeros of Quadratic: Substitute $a$ , $b$ , and $c$ into the quadratic formula to find the zeros of $x^2 + 6x + 2$ . $\newline$ Calculation: $x = \frac{-6 \pm \sqrt{6^2 - 4\cdot1\cdot2}}{2\cdot1} = \frac{-6 \pm \sqrt{36 - 8}}{2} = \frac{-6 \pm \sqrt{28}}{2} = \frac{-6 \pm 2\sqrt{7}}{2} = -3 \pm \sqrt{7}$
Express Quadratic as Product: The zeros of the quadratic are $-3 + \sqrt{7}$ and $-3 - \sqrt{7}$ , so the quadratic can be written as $(x - (-3 + \sqrt{7}))(x - (-3 - \sqrt{7}))$ , which simplifies to $(x + 3 - \sqrt{7})(x + 3 + \sqrt{7})$ .
Final Expression of $h(x)$ : Now we can express $h(x)$ as a product of linear factors: $h(x) = (x - 3)(x + 3 - \sqrt{7})(x + 3 + \sqrt{7})$ .