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$x^3 + 2x^2- 9x -18 = (x^2 - a^2)(x+b)$ were $a$ and $b$ are integers. Work out the three linear factors of $x^3 + 2x^2 - 9x - 18$. What would I do to answer this? I don't have a clue where to start

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$x^3+2x^2-9x-18 = x^2(x+2)-9(x+2)=(x+2)(x^2-9)=(x+2)(x-3)(x+3)$

Take the common factor from the first pair, $x^3+2x^2=x^2(x+2)$, and second pair of terms, $-9x-18=-9(x+2)$. Notice that this has a common factor $(x+2)$ and can be factored out next. Then, taking a difference of squares, $(x^2-a^2)=(x+a)(x-a)$ completes the factoring. This is coming at it from a different route.

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Just multiply out the expression on the right and equate coefficients of the same powers of $x$ on both sides. You immediately get values for $b$ and $a^2$. You should know how to factor $x^2-a^2$, right?

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I'm still really stuck on it.. – H.cowper Apr 4 '14 at 21:53
Do the polynomial multiplication on the right side: $(x^2-a^2)(x+b)$ can be multiplied out to give $x^2\cdot(x+b) - a^2\cdot(x+b)$. Then continue simplifying to get $x^3+bx^2-a^2x -a^2b$, right? Now compare this to the left side and you can read off what $a$ and $b$ are. – MPW Apr 4 '14 at 21:57


\begin{equation*} f(x)=x^{3}+2x^{2}-9x-18 \end{equation*}

can be written as

\begin{equation*}f(x)=\left( x^{2}-a^{2}\right) (x+b), \end{equation*}

then expanding the right hand side yields

\begin{equation*} f(x)=x^{3}+bx^{2}-a^{2}x-a^{2}b. \end{equation*}

To ensure that for all $x$


the coefficients of both sides must be equal, thus finding that

\begin{equation*} b=2 \text{ (coeff. of }x^2),\qquad -a^{2}=-9 \text{ (coeff. of }x),\qquad -a^{2}b=-18 \text{ (coeff. of }x^0). \end{equation*}

So $a^2=9,b=2$ and

\begin{equation*} f(x)=x^{3}+2x^{2}-9x-18=\left( x^{2}-a^{2}\right) (x+b)=(x^{2}-9)(x+2)=(x-3)(x+3)(x+2), \end{equation*}

where $(x-3)$, $(x+3)$ and $(x+2)$ are the three linear factors asked in the question.

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