Question
Use synthetic division to decompose the following polynomial into two factors knowing that (x + 2) is a factor: 10x3+11x2-27x-18
Answers
GPT-4o mini
To decompose the polynomial \(10x^3 + 11x^2 - 27x - 18\) using synthetic division by the factor \((x + 2)\), we first set \(x + 2 = 0\), which gives us \(x = -2\).
Next, we can perform synthetic division using \(-2\), and the coefficients of the polynomial \(10, 11, -27, -18\):
1. Write down the coefficients:
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & & & \\
\end{array}
\]
2. Bring down the leading coefficient (10):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & & & \\
& 10 & & & \\
\end{array}
\]
3. Multiply \(-2\) by \(10\) and add to the next coefficient (11):
\(-2 \times 10 = -20\), and \(11 + (-20) = -9\):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & -20 & & \\
& 10 & -9 & & \\
\end{array}
\]
4. Multiply \(-2\) by \(-9\) and add to \(-27\):
\(-2 \times -9 = 18\), and \(-27 + 18 = -9\):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & -20 & 18 & \\
& 10 & -9 & -9 & \\
\end{array}
\]
5. Multiply \(-2\) by \(-9\) and add to \(-18\):
\(-2 \times -9 = 18\), and \(-18 + 18 = 0\):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & -20 & 18 & 0 \\
& 10 & -9 & -9 & 0 \\
\end{array}
\]
The result of the synthetic division is:
\[
10x^2 - 9x - 9
\]
Since the remainder is \(0\), we can confirm that \(x + 2\) is a factor of the original polynomial.
Next, we can factor the quadratic \(10x^2 - 9x - 9\). To factor this quadratic, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 10\), \(b = -9\), and \(c = -9\).
Calculating the discriminant:
\[
b^2 - 4ac = (-9)^2 - 4 \cdot 10 \cdot (-9) = 81 + 360 = 441
\]
Now calculating the roots:
\[
x = \frac{9 \pm \sqrt{441}}{20} = \frac{9 \pm 21}{20}
\]
Calculating the two roots:
1. \(x = \frac{30}{20} = \frac{3}{2}\)
2. \(x = \frac{-12}{20} = -\frac{3}{5}\)
Thus, we can factor the quadratic as:
\[
10x^2 - 9x - 9 = 10(x - \frac{3}{2})(x + \frac{3}{5})
\]
However, for simplicity, we can also write it as:
\[
10(x - \frac{3}{2})(x + \frac{3}{5}) = (x + 2)(10x^2 - 9x - 9)
\]
The complete factorization can be written as:
\[
10(x + 2)\left(x - \frac{3}{2}\right)\left(x + \frac{3}{5}\right)
\]
Summarizing, the factors of \(10x^3 + 11x^2 - 27x - 18\) are:
\[
(x + 2)(10x^2 - 9x - 9)
\]
Next, we can perform synthetic division using \(-2\), and the coefficients of the polynomial \(10, 11, -27, -18\):
1. Write down the coefficients:
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & & & \\
\end{array}
\]
2. Bring down the leading coefficient (10):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & & & \\
& 10 & & & \\
\end{array}
\]
3. Multiply \(-2\) by \(10\) and add to the next coefficient (11):
\(-2 \times 10 = -20\), and \(11 + (-20) = -9\):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & -20 & & \\
& 10 & -9 & & \\
\end{array}
\]
4. Multiply \(-2\) by \(-9\) and add to \(-27\):
\(-2 \times -9 = 18\), and \(-27 + 18 = -9\):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & -20 & 18 & \\
& 10 & -9 & -9 & \\
\end{array}
\]
5. Multiply \(-2\) by \(-9\) and add to \(-18\):
\(-2 \times -9 = 18\), and \(-18 + 18 = 0\):
\[
\begin{array}{r|rrrr}
-2 & 10 & 11 & -27 & -18 \\
& & -20 & 18 & 0 \\
& 10 & -9 & -9 & 0 \\
\end{array}
\]
The result of the synthetic division is:
\[
10x^2 - 9x - 9
\]
Since the remainder is \(0\), we can confirm that \(x + 2\) is a factor of the original polynomial.
Next, we can factor the quadratic \(10x^2 - 9x - 9\). To factor this quadratic, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 10\), \(b = -9\), and \(c = -9\).
Calculating the discriminant:
\[
b^2 - 4ac = (-9)^2 - 4 \cdot 10 \cdot (-9) = 81 + 360 = 441
\]
Now calculating the roots:
\[
x = \frac{9 \pm \sqrt{441}}{20} = \frac{9 \pm 21}{20}
\]
Calculating the two roots:
1. \(x = \frac{30}{20} = \frac{3}{2}\)
2. \(x = \frac{-12}{20} = -\frac{3}{5}\)
Thus, we can factor the quadratic as:
\[
10x^2 - 9x - 9 = 10(x - \frac{3}{2})(x + \frac{3}{5})
\]
However, for simplicity, we can also write it as:
\[
10(x - \frac{3}{2})(x + \frac{3}{5}) = (x + 2)(10x^2 - 9x - 9)
\]
The complete factorization can be written as:
\[
10(x + 2)\left(x - \frac{3}{2}\right)\left(x + \frac{3}{5}\right)
\]
Summarizing, the factors of \(10x^3 + 11x^2 - 27x - 18\) are:
\[
(x + 2)(10x^2 - 9x - 9)
\]