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

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!!

MickieMik said:

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!!

Click to expand...

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.

MickieMik said:

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.

Click to expand...

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.

MickieMik said:

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!!

Click to expand...

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

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

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)