Factoring Trinomials: 3x^2 + 23x - 36

[MickieMik]

OK, the last one I answered myself, hopefully, this one I've spent way too much time on and really need help.
Factor Completely
3x^2 + 23x - 36
Nothing I could come up with would work so in the book it shows you how it find out if a trinomial is factorable. The formula they give is ax^1 +bx +c which would be-
a=3, b=23, and c=36 then use the formula b^2 - 4ac to see if it is factorable
which would be - 23^2 - 4(3)(36) which would give
529 - (-432) which gives 961 which is a perfect square (31^2)
OK, but then it says once you figure that out our go back to the regular way to factor a trinomial but nothing factors because 3 does not go into 23 and nothing that I could find that factors the other numbers will equal 23. Where am I going wrong?
Please Help!! Thank you so very much!!

 

[Mrspi]

OK, the last one I answered myself, hopefully, this one I've spent way too much time on and really need help.
Factor Completely
3x^2 + 23x - 36
Nothing I could come up with would work so in the book it shows you how it find out if a trinomial is factorable. The formula they give is ax^1 +bx +c which would be-
a=3, b=23, and c=36 then use the formula b^2 - 4ac to see if it is factorable
which would be - 23^2 - 4(3)(36) which would give
529 - (-432) which gives 961 which is a perfect square (31^2)
OK, but then it says once you figure that out our go back to the regular way to factor a trinomial but nothing factors because 3 does not go into 23 and nothing that I could find that factors the other numbers will equal 23. Where am I going wrong?
Please Help!! Thank you so very much!!

There is no common factor greater than 1 which divides all three terms. So, you're "stuck" with factoring 3x^2 + 23x - 36

Here's how I do these. Multiply the coefficient of x^2 by the constant term. 3 * (-36) is -108

Now, see if you can find two numbers which multiply to -108, and which add up to 23 (the coefficient of the middle term). This may take a bit of trying, but I found that -4 and +27 work, since -4(27) = -108, and -4 + 27 = 23.

Rewrite the middle term as -4x + 27x:

3x^2 - 4x + 27x - 36

Now, group the first two terms together, and the last two terms together. Remove the greatest common factor from each pair of terms.

The first two terms have a common factor of x. The last two terms have a common factor of 9.

x(3x - 4) + 9(3x - 4)

NOW....(3x - 4) is a common factor. Remove it, and you have

(3x - 4)(x + 9)

You can verify that this is the correct factorization by multiplying the two binomials together. You should get 3x^2 + 23x - 36.

 

[stapel]

Factor Completely: 3x^2 + 23x - 36
Nothing I could come up with would work so in the book it shows you how it find out if a trinomial is factorable....
...nothing factors because 3 does not go into 23 and nothing that I could find that factors the other numbers will equal 23.

So you know the trinomial is factorable, but have not been taught how to factor, other than taking a common factor out of all three terms...?

Have you done any multiplication of binomials at all? For instance, would you know how to multiply factors like (x + 2) and (4x - 5)? If so, then one method for factoring trinomials of this sort is to factor "a" and "c" (3 and -36, respectively, in this case), plug the factorizations into parentheses, and see what works. Since 3 factors as 1 and 3, and since 36 factors as 1 and 36, 2 and 18, 3 and 12, 4 and 9, and 6 and 6, you'd have (with the signs included) the following factor pairs to test:

. . . . .(x + 1)(3x - 36)
. . . . .(x - 1)(3x + 36)
. . . . .(x + 36)(3x - 1)
. . . . .(x - 36)(3x + 1)
. . . . .(x + 2)(3x - 18)
. . . . .(x - 2)(3x + 28)
. . . . .(x + 18)(3x - 1)
. . . . .(x - 18)(3x + 1)
. . . . .(x + 3)(3x - 12)
. . . . .(x - 3)(3x + 12)
. . . . .(x + 12)(3x - 3)
. . . . .(x - 12)(3x + 3)
. . . . .(x + 4)(3x - 9)
. . . . .(x - 4)(3x + 9)
. . . . .(x + 9)(3x - 4)
. . . . .(x - 9)(3x + 4)
. . . . .(x + 6)(3x - 6)
. . . . .(x - 6)(3x + 6)

(I think that's all of them.) Multiply them out until you find one that gives you 3x[sup:3521e696]2[/sup:3521e696] + 23x - 36.

There are methods for factoring which can simplify this process somewhat, but I don't want to overwhelm you. If you've done multiplying at all, the above will get the job done. :wink:

Eliz.

 

[Deleted member 4993]

OK, the last one I answered myself, hopefully, this one I've spent way too much time on and really need help.
Factor Completely
3x^2 + 23x - 36
Nothing I could come up with would work so in the book it shows you how it find out if a trinomial is factorable. The formula they give is ax^1 +bx +c which would be-
a=3, b=23, and c=36 then use the formula b^2 - 4ac to see if it is factorable
which would be - 23^2 - 4(3)(36) which would give
529 - (-432) which gives 961 which is a perfect square (31^2)
OK, but then it says once you figure that out our go back to the regular way to factor a trinomial but nothing factors because 3 does not go into 23 and nothing that I could find that factors the other numbers will equal 23. Where am I going wrong?
Please Help!! Thank you so very much!!

\(\displaystyle A\cdot x^2\, + \,B\cdot x \, + \, C\)

\(\displaystyle =\, A\cdot [x\, - \, x_1]\cdot [x\, - \, x_2]........................(1)\)

where

\(\displaystyle x_1 \, =\, \frac{-B\, + \,\sqrt{B^2\, - \,4\cdot A \cdot C}}{2\cdot A}\)

and


\(\displaystyle x_2 \, =\, \frac{-B\, - \,\sqrt{B^2\, - \,4\cdot A \cdot C}}{2\cdot A}\)

calculate

\(\displaystyle x_1\, and \, x_2\)

and use in equation (1)