Question
what are the linear factors of 18x^3+63x^2-8x-28
Answers
Answered by
GPT-4o mini
To factor the polynomial \(18x^3 + 63x^2 - 8x - 28\), we can use the Rational Root Theorem to test possible rational roots, and then use synthetic division or polynomial long division to factor the polynomial.
First, let's rewrite the polynomial:
\[
f(x) = 18x^3 + 63x^2 - 8x - 28
\]
The Rational Root Theorem states that any rational root, in simplest form \(\frac{p}{q}\), must have \(p\) as a factor of the constant term (-28) and \(q\) as a factor of the leading coefficient (18).
Factors of -28 are \( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \).
Factors of 18 are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
Thus, the possible rational roots are:
\[
\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{1}{9}, \pm \frac{1}{18}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 4, \pm 7, \pm 14, \pm 28
\]
Next, we can test these values in \(f(x)\).
After testing several possible roots, for example:
- \(f(1) = 18(1)^3 + 63(1)^2 - 8(1) - 28 = 45 \) (not a root)
- \(f(-1) = 18(-1)^3 + 63(-1)^2 - 8(-1) - 28 = 21 \) (not a root)
- \(f(2) = 18(2)^3 + 63(2)^2 - 8(2) - 28 = 0 \) (is a root)
Since \(x = 2\) is a root, we can divide the polynomial by \(x - 2\).
Using synthetic division:
\[
\begin{array}{r|rrrr}
2 & 18 & 63 & -8 & -28 \\
& & 36 & 198 & 380 \\
\hline
& 18 & 99 & 190 & 0 \\
\end{array}
\]
The quotient after the division is \(18x^2 + 99x + 190\).
Next, we need to factor \(18x^2 + 99x + 190\). We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 18\), \(b = 99\), and \(c = 190\):
\[
b^2 - 4ac = 99^2 - 4 \cdot 18 \cdot 190 = 9801 - 13680 = -3789
\]
Since the discriminant is negative, \(18x^2 + 99x + 190\) does not have real roots, meaning it cannot be factored further over the reals.
Thus, the linear factor we found is \( (x - 2) \), and the remainder is the irreducible quadratic \( 18x^2 + 99x + 190\).
Finally, the complete factorization of the polynomial is:
\[
f(x) = (x - 2)(18x^2 + 99x + 190)
\]
So, the linear factor of \(18x^3 + 63x^2 - 8x - 28\) is \(x - 2\).
First, let's rewrite the polynomial:
\[
f(x) = 18x^3 + 63x^2 - 8x - 28
\]
The Rational Root Theorem states that any rational root, in simplest form \(\frac{p}{q}\), must have \(p\) as a factor of the constant term (-28) and \(q\) as a factor of the leading coefficient (18).
Factors of -28 are \( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \).
Factors of 18 are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
Thus, the possible rational roots are:
\[
\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{1}{9}, \pm \frac{1}{18}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 4, \pm 7, \pm 14, \pm 28
\]
Next, we can test these values in \(f(x)\).
After testing several possible roots, for example:
- \(f(1) = 18(1)^3 + 63(1)^2 - 8(1) - 28 = 45 \) (not a root)
- \(f(-1) = 18(-1)^3 + 63(-1)^2 - 8(-1) - 28 = 21 \) (not a root)
- \(f(2) = 18(2)^3 + 63(2)^2 - 8(2) - 28 = 0 \) (is a root)
Since \(x = 2\) is a root, we can divide the polynomial by \(x - 2\).
Using synthetic division:
\[
\begin{array}{r|rrrr}
2 & 18 & 63 & -8 & -28 \\
& & 36 & 198 & 380 \\
\hline
& 18 & 99 & 190 & 0 \\
\end{array}
\]
The quotient after the division is \(18x^2 + 99x + 190\).
Next, we need to factor \(18x^2 + 99x + 190\). We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 18\), \(b = 99\), and \(c = 190\):
\[
b^2 - 4ac = 99^2 - 4 \cdot 18 \cdot 190 = 9801 - 13680 = -3789
\]
Since the discriminant is negative, \(18x^2 + 99x + 190\) does not have real roots, meaning it cannot be factored further over the reals.
Thus, the linear factor we found is \( (x - 2) \), and the remainder is the irreducible quadratic \( 18x^2 + 99x + 190\).
Finally, the complete factorization of the polynomial is:
\[
f(x) = (x - 2)(18x^2 + 99x + 190)
\]
So, the linear factor of \(18x^3 + 63x^2 - 8x - 28\) is \(x - 2\).