Question
Use synthetic division to show that 3 is a solution of the polynomial equation below.
10 x cubed minus 15 x squared minus 15 x minus 90equals0
b. Use the solution from part (a) to solve this problem. The number of eggs, f(x), in a female moth is a function of her abdominal width, in millimeters, modeled by the equation below.
f(x)equals10 x cubed minus 15 x squared minus 15 x minus 6
What is the abdominal width when there are 84 eggs?
Question content area bottom
Part 1
a. The number 3 is a solution to the equation because the remainder of the division, 10 x cubed minus 15 x squared minus 15 x minus 90 divided by xminus3, is
enter your response here.
Part 2
b. The abdominal width is
enter your response here millimeters.
10 x cubed minus 15 x squared minus 15 x minus 90equals0
b. Use the solution from part (a) to solve this problem. The number of eggs, f(x), in a female moth is a function of her abdominal width, in millimeters, modeled by the equation below.
f(x)equals10 x cubed minus 15 x squared minus 15 x minus 6
What is the abdominal width when there are 84 eggs?
Question content area bottom
Part 1
a. The number 3 is a solution to the equation because the remainder of the division, 10 x cubed minus 15 x squared minus 15 x minus 90 divided by xminus3, is
enter your response here.
Part 2
b. The abdominal width is
enter your response here millimeters.
Answers
GPT-4o mini
To show that \( x = 3 \) is a solution of the polynomial equation \( 10x^3 - 15x^2 - 15x - 90 = 0 \) using synthetic division, we will perform synthetic division of the polynomial \( 10x^3 - 15x^2 - 15x - 90 \) by \( x - 3 \).
**Part 1: Synthetic Division**
1. Write down the coefficients of the polynomial:
\( 10, -15, -15, -90 \)
2. Set up synthetic division with \( 3 \):
\[
\begin{array}{r|rrrr}
3 & 10 & -15 & -15 & -90 \\
& & 30 & 45 & 90 \\
\hline
& 10 & 15 & 30 & 0 \\
\end{array}
\]
3. Perform the synthetic division:
- Bring down the \( 10 \).
- Multiply \( 3 \times 10 = 30 \) and add it to \(-15\) to get \( 15 \).
- Multiply \( 3 \times 15 = 45 \) and add it to \(-15\) to get \( 30 \).
- Multiply \( 3 \times 30 = 90 \) and add it to \(-90\) to get \( 0 \).
The remainder is \( 0 \), which means \( 3 \) is indeed a root of the polynomial \( 10x^3 - 15x^2 - 15x - 90 \).
Thus, the answer to the first part is:
\[
\text{The number 3 is a solution to the equation because the remainder of the division } 10x^3 - 15x^2 - 15x - 90 \text{ divided by } x - 3 \text{ is } 0.
\]
**Part 2: Solving for Abdominal Width**
Now we use the solution from part (a) to find the abdominal width when there are \( 84 \) eggs. We need to solve the equation:
\[
f(x) = 10x^3 - 15x^2 - 15x - 6 = 84
\]
Rearranging gives us:
\[
10x^3 - 15x^2 - 15x - 6 - 84 = 0
\]
which simplifies to:
\[
10x^3 - 15x^2 - 15x - 90 = 0
\]
Now we can apply synthetic division again knowing that \( x = 3 \) is a root:
Since we already know that \( 10x^3 - 15x^2 - 15x - 90 = (x-3)(10x^2 + 15x + 30) \).
Now, we can find the roots of \( 10x^2 + 15x + 30 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
with \( a = 10, b = 15, c = 30 \):
Calculating the discriminant:
\[
b^2 - 4ac = 15^2 - 4(10)(30) = 225 - 1200 = -975
\]
Since the discriminant is negative, the quadratic \( 10x^2 + 15x + 30 \) does not have real roots. Therefore, the only real solution for the number of eggs \( f(x) = 84 \) occurs at \( x = 3 \).
Thus, the abdominal width when there are \( 84 \) eggs is:
\[
\text{The abdominal width is } 3 \text{ millimeters.}
\]
**Part 1: Synthetic Division**
1. Write down the coefficients of the polynomial:
\( 10, -15, -15, -90 \)
2. Set up synthetic division with \( 3 \):
\[
\begin{array}{r|rrrr}
3 & 10 & -15 & -15 & -90 \\
& & 30 & 45 & 90 \\
\hline
& 10 & 15 & 30 & 0 \\
\end{array}
\]
3. Perform the synthetic division:
- Bring down the \( 10 \).
- Multiply \( 3 \times 10 = 30 \) and add it to \(-15\) to get \( 15 \).
- Multiply \( 3 \times 15 = 45 \) and add it to \(-15\) to get \( 30 \).
- Multiply \( 3 \times 30 = 90 \) and add it to \(-90\) to get \( 0 \).
The remainder is \( 0 \), which means \( 3 \) is indeed a root of the polynomial \( 10x^3 - 15x^2 - 15x - 90 \).
Thus, the answer to the first part is:
\[
\text{The number 3 is a solution to the equation because the remainder of the division } 10x^3 - 15x^2 - 15x - 90 \text{ divided by } x - 3 \text{ is } 0.
\]
**Part 2: Solving for Abdominal Width**
Now we use the solution from part (a) to find the abdominal width when there are \( 84 \) eggs. We need to solve the equation:
\[
f(x) = 10x^3 - 15x^2 - 15x - 6 = 84
\]
Rearranging gives us:
\[
10x^3 - 15x^2 - 15x - 6 - 84 = 0
\]
which simplifies to:
\[
10x^3 - 15x^2 - 15x - 90 = 0
\]
Now we can apply synthetic division again knowing that \( x = 3 \) is a root:
Since we already know that \( 10x^3 - 15x^2 - 15x - 90 = (x-3)(10x^2 + 15x + 30) \).
Now, we can find the roots of \( 10x^2 + 15x + 30 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
with \( a = 10, b = 15, c = 30 \):
Calculating the discriminant:
\[
b^2 - 4ac = 15^2 - 4(10)(30) = 225 - 1200 = -975
\]
Since the discriminant is negative, the quadratic \( 10x^2 + 15x + 30 \) does not have real roots. Therefore, the only real solution for the number of eggs \( f(x) = 84 \) occurs at \( x = 3 \).
Thus, the abdominal width when there are \( 84 \) eggs is:
\[
\text{The abdominal width is } 3 \text{ millimeters.}
\]