Any pointers for factoring trinomial into two binomials?

[samalex]

I know that a^2+2ab+b^2 = (a+b)^2 but I'm having a hard time coming up with a quick or simple way to factor trinomials like this
7x^2 - 16x - 15 into this (7x+5)(x-3).

Several examples I've read say to get the two numbers who's product would be -15 and sum would be -16, but though the product of +5 and -3 is -15 their sum is not -16. I can use FOIL on the two binomials to see this is correct, but is there any quick or simple way other than trial and error to factor a trinomial like this to two binomials?

Thanks for any pointers ...

Sam

 

[Mrspi]

I know that a^2+2ab+b^2 = (a+b)^2 but I'm having a hard time coming up with a quick or simple way to factor trinomials like this
7x^2 - 16x - 15 into this (7x+5)(x-3).

Several examples I've read say to get the two numbers who's product would be -15 and sum would be -16, but though the product of +5 and -3 is -15 their sum is not -16. I can use FOIL on the two binomials to see this is correct, but is there any quick or simple way other than trial and error to factor a trinomial like this to two binomials?

Thanks for any pointers ...

Sam

Here's what is sometimes called the "ac method".....works nicely when the coefficient of the SQUARED term is something other than 1.

I'll use your example problem to explain how it works:

7x^2 - 16x - 15

"a" in this case is 7, "c" is -15, and "b" is -16.

1) multiply a*c. 7*(-15) = -105

2) find two factors of the product ac which have a sum that's equal to "b". In other words, we need to find two factors of -105 which sum to -16. We can use the following pairs of factors to get 105: 1*105, 3*35, 5*21, and 7*15. To get a NEGATIVE 105, one factor in the pair must be negative. A little examination shows us that -21*5 is the pair we want, since -21 + 5 = -16. Note...if there is no pair of factors of "ac" which add up to "b", then the expression will not factor over the integers.

3) use the pair of numbers found in step 2 to re-write the middle term. Since we know that -21 + 5 = -16, we can rewrite the middle term as -21x + 5x:

7x^2 - 21x + 5x - 15

4) Factor by grouping the first two terms together, and the last two terms together, and removing a common factor from each group:

(7x^2 - 21x) + (5x - 15)
7x(x - 3) + 5(x - 3)
(x - 3)(7x + 5)

I have used this method for years, and find it very effective. Of course, opinions may vary....

 

[jsterkel]

Mrspi,

That is a really good explanation.

Justin

 

[lookagain]

. . .
7x^2 - 16x - 15 into this (7x+5)(x-3).

samalex,

if you want to even know whether to bother looking for
the possible factored form of a trinomial, ax^2 + bx + c,
(where all of those coefficents are integers and "a" is not
equal to 0,then figure out the value of the discriminant,
b^2 - 4ac, first.


\(\displaystyle If \ b^2 - 4ac \ does \ not \ equal \ to \ 0 \ or \ a\)

\(\displaystyle positive \ perfect-square \ integer, \ then \ the \ \)

\(\displaystyle trinomial \ will \ not \ factor \ (over \ the \ integers).\)

 

[Mrspi]

samalex,

if you want to even know whether to bother looking for
the possible factored form of a trinomial, ax^2 + bx + c,
(where all of those coefficents are integers and "a" is not
equal to 0,then figure out the value of the discriminant,
b^2 - 4ac, first.


\(\displaystyle If \ b^2 - 4ac \ does \ not \ equal \ to \ 0 \ or \ a\)

\(\displaystyle positive \ perfect-square \ integer, \ then \ the \ \)

\(\displaystyle trinomial \ will \ not \ factor \ (over \ the \ integers).\)

That's an excellent point, Lookagain!

 

[soroban]

Hello, samalex!

I prefer a slightly different version of the "ac method".
It ignores the signs of the factors until the end.

See what you think . . .

\(\displaystyle \text{Factor: }\:7x^2 - 16x - 15\)

[1] Note the sign of the constant term.
. . .If plus, think "sum".
. . .If minus, think "difference".

. . .The constant term has minus.
. . .Remember "difference" for step 3.


[2] Multiply the leading coefficient by the constant.
. . .We have: .\(\displaystyle ac \:=\:7\cdot15 \:=\:105\)


[3] Factor \(\displaystyle ac\) into two factors
. . .whose sum/difference is the middle coefficient.

. . .Factor 105 into two parts whose difference is 16.

. . .We factor 105 by dividing it by 1, 2, 3, ...
. . . and keep the ones that "come out even.

. . \(\displaystyle \begin{array}{cc}\text{Factors} & \text{Diff.}\\ \hline 1,105 & 104\\ 2,?\;\; & -- \\ 3, 35 & 32 \\ 4, ?\;\; & -- \\ 5, 21 & 16 \\ \vdots & \vdots \end{array}\begin{array}{c} \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}\)


. . The pair of factors is: .\(\displaystyle (5,21)\)

. . The middle term has -16, so we will use +5 and -21.


\(\displaystyle \text{[4] Replace -}16x\text{ with }+5x\text{ and -}21x.\)

. . .. . \(\displaystyle 7x^2 + 5x - 21x - 15\)


[5] Factor by "grouping".

. . .\(\displaystyle 7x^2 + 5x - 21x - 15\)

. . . . \(\displaystyle =\;x(7x + 5) - 3(7x + 5)\)

. . . . \(\displaystyle =\;(7x+5)(x-3)\)