Question

question\
ewline $3x^2 + 25x - 18$\
ewline which of the following is a factor of the polynomial above?\
ewline $\circ$ $x - 9$\
ewline $\circ$ $x + 3$\
ewline $\circ$ $3x - 2$\
ewline $\circ$ $3x + 1$

Explanation:

Step1: Factor the quadratic polynomial

We need to factor \(3x^2 + 25x - 18\). We look for two numbers that multiply to \(3\times(-18)= - 54\) and add up to \(25\). The numbers are \(27\) and \(- 2\) since \(27\times(-2)=-54\) and \(27+( - 2)=25\).
Rewrite the middle term: \(3x^2+27x - 2x - 18\)
Group the terms: \((3x^2 + 27x)+(-2x - 18)\)
Factor out the common factors from each group: \(3x(x + 9)-2(x + 9)\)
Now factor out \((x + 9)\): \((3x - 2)(x + 9)\) Wait, no, wait, let's check again. Wait, maybe I made a mistake. Wait, let's try another way. Wait, the quadratic is \(3x^2+25x - 18\). Let's use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 25\), \(c=-18\).
\(x=\frac{-25\pm\sqrt{25^2-4\times3\times(-18)}}{2\times3}=\frac{-25\pm\sqrt{625 + 216}}{6}=\frac{-25\pm\sqrt{841}}{6}=\frac{-25\pm29}{6}\)
So \(x=\frac{-25 + 29}{6}=\frac{4}{6}=\frac{2}{3}\) and \(x=\frac{-25-29}{6}=\frac{-54}{6}=-9\)
So the factors are \((x + 9)\) and \((3x - 2)\) (since if \(x=\frac{2}{3}\), then \(3x-2 = 0\)). Wait, but the options are \(x - 9\), \(x + 3\), \(3x-2\), \(3x + 1\). So from the factors we found, \(3x - 2\) is a factor. Let's verify by multiplying \((3x - 2)(x + 9)=3x^2+27x-2x - 18=3x^2+25x - 18\), which matches the original polynomial.

Step2: Check the options

We can also use the factor theorem. For a polynomial \(f(x)\), if \(f(a)=0\), then \((x - a)\) is a factor. Let's test each option:

• For option \(x - 9\), \(f(9)=3\times9^2+25\times9 - 18=3\times81 + 225-18=243+225 - 18=450

eq0\)

• For option \(x + 3\), \(f(-3)=3\times(-3)^2+25\times(-3)-18=27-75 - 18=-66

eq0\)

• For option \(3x - 2\), let \(3x-2 = 0\), \(x=\frac{2}{3}\). \(f(\frac{2}{3})=3\times(\frac{2}{3})^2+25\times\frac{2}{3}-18=3\times\frac{4}{9}+\frac{50}{3}-18=\frac{4}{3}+\frac{50}{3}-18=\frac{54}{3}-18 = 18 - 18=0\). So \(3x - 2\) is a factor.
• For option \(3x + 1\), let \(3x+1 = 0\), \(x=-\frac{1}{3}\). \(f(-\frac{1}{3})=3\times(-\frac{1}{3})^2+25\times(-\frac{1}{3})-18=3\times\frac{1}{9}-\frac{25}{3}-18=\frac{1}{3}-\frac{25}{3}-18=\frac{-24}{3}-18=-8 - 18=-26

eq0\)

Answer:

\(3x - 2\) (the option corresponding to \(3x - 2\))