1.5 Factoring trinomials with leading coefficient other than 1  (Page 2/3)

Factor $8x^2-30x-27$ .

Find the factors of the first and last terms.



Thus, the $4x$ and $-9$ are to be multiplied, and $2x$ and 3 are to be multiplied.

$\begin{array}{lll}8x^2-30x-27\hfill & =\hfill & (\begin{array}{c}4x\end{array})(\begin{array}{c}2x\end{array})\hfill \\ \hfill & =\hfill & (\begin{array}{c}4x\end{array})(\begin{array}{ccc}2x& -& 9\end{array})\hfill \\ \hfill & =\hfill & (\begin{array}{ccc}4x& +& 3\end{array})(\begin{array}{ccc}2x& -& 9\end{array})\hfill \end{array}$

$\begin{array}{lll}check:(4x+3)(2x-9)\hfill & =\hfill & 8x^2-36x+6x-27\hfill \\ \hfill & =\hfill & 8x^2-30x-27\hfill \end{array}$

Factor $15x^2+44x+32$ .

Before we start finding the factors of the first and last terms, notice that the constant term is $+32$ . Since the product is positive , the two factors we are looking for must have the same sign. They must both be positive or both be negative. Now the middle term, $+44x$ , is preceded by a positive sign. We know that the middle term comes from the sum of the outer and inner products. If these two numbers are to sum to a positive number, they must both be positive themselves. If they were negative, their sum would be negative. Thus, we can conclude that the two factors of $+32$ that we are looking for are both positive numbers. This eliminates several factors of 32 and lessens our amount of work.
Factor the first and last terms.



After a few trials we see that $5x$ and 4 are to be multiplied, and $3x$ and 8 are to be multiplied.

$15x^2+44x+32=(5x+8)(3x+4)$

Factor $18x^2-56x+6$ .

We see that each term is even, so we can factor out 2.

$2(9x^2-28x+3)$

Notice that the constant term is positive. Thus, we know that the factors of 3 that we are looking for must have the same sign. Since the sign of the middle term is negative, both factors must be negative.
Factor the first and last terms.



There are not many combinations to try, and we find that $9x$ and $-3$ are to be multiplied and $x$ and $-1$ are to be multiplied.

$\begin{array}{lll}18x^2-56x+6\hfill & =\hfill & 2(9x^2-28x+3)\hfill \\ \hfill & =\hfill & 2(9x-1)(x-3)\hfill \end{array}$

If we had not factored the 2 out first, we would have gotten the factorization

The factorization is not complete since one of the factors may be factored further.

$\begin{array}{llll}18x^2-56x+6\hfill & =\hfill & (9x-1)(2x-6)\hfill & \hfill \\ \hfill & =\hfill & (9x-1)\cdot 2(x-3)\hfill & \hfill \\ \hfill & =\hfill & 2(9x-1)(x-3)\hfill & (\text{By}\text{the}\text{commutative}\text{property}\text{of}\text{multiplication}\text{.})\hfill \end{array}$

The results are the same, but it is much easier to factor a polynomial after all common factors have been factored out first.

Factor $3x^2+x-14$ .

There are no common factors. We see that the constant term is negative. Thus, the factors of $-14$ must have different signs.
Factor the first and last terms.



After a few trials, we see that $3x$ and $-2$ are to be multiplied and $x$ and 7 are to be multiplied.

$3x^2+x-14=(3x+7)(x-2)$

Factor $8x^2-26xy+15y^2$ .

We see that the constant term is positive and that the middle term is preceded by a minus sign.
Hence, the factors of $15y^2$ that we are looking for must both be negative.
Factor the first and last terms.



After a few trials, we see that $4x$ and $-5y$ are to be multiplied and $2x$ and $-3y$ are to be multiplied.

$8x^2-26xy+15y^2=(4x-3y)(2x-5y)$