So we have
. Lets take a guess and check if
)
will give us zero meaning
is a factor.
 & =16-36-4+24<br />\\ & =20-20<br />\\ & =0<br />\\ \therefore x-2\text{ is a factor}<br />\end{alignedat})
So now we know that we can express our expression in the form
\left(ax^{2}+bx+c\right))
(since the left over factor has to be one less than the original as we took a linear factor out).
\left(ax^{2}+bx+c\right)\end{alignedat})
Now we need to find each of the coefficients, so we'll look at what happens if we expand out the RHS and then equate coefficients on each of the terms we have.
So, the only way we can get an
term is from the multiplication of the
and
terms together. That is
This coefficient has to be equal to the coefficient on the
on the left hand side, so
.
Now for the
terms, how can we get
terms when we expand the RHS above? From multiplying together ex]x^{2}[/tex] with
terms and from multiplying together
terms with other
terms.
+\left(-2\times ax^{2}\right) & =bx^{2}-2ax^{2}<br />\\ & =\left(-2a+b\right)x^{2}<br />\end{alignedat})
Which again, we know the coefficient has to be equal to the coefficient on the
on the LHS.
Similarily we have the
terms (from multiplying together tex]x^{1}[/tex] and
terms). We can also do the same for the
terms as this gives us
without the need to substitute it back into anything else.
 & =24<br />\\ \implies c & =-12<br />\end{alignedat})
So then writing the original expression in it's new form gives
 & =\left(x-2\right)\left(2x^{2}-5x-12\right)<br />\\ \therefore P\left(x\right) & =\left(x-2\right)\left(2x+3\right)\left(x-4\right)<br />\end{alignedat})
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