I have a doubt with the following problem I found in a book. You have to simplify a polynomial using the GCF. Now, this is the problem I am not able to grasp:
$$6x^2-19x-7$$
According to the book, the solution is
$$(2x-7)(3x+1)$$
And, in fact, if you multiply those two terms you'll get:
$$6x^2-21x+2x-7=6x^2-19x-7$$
So it is clear that both expressions are equivalent. What I fail to grasp is which method they have followed to come from $6x^2-19x-7$ to $(2x-7)(3x+1)$.
Thanks in advance,
On
So the way I will do it is like this.
First, check if the equation $6x^2−19x−7 = 0 $ has solutions by calculating
$b^2 - 4ac = 19^2 + 4 * 6 * 7 = 529$
Then, we can get the roots from the formula $x = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a} = \frac{19 +/- \sqrt{529}}{2* 6}$ = 3.5/ $-\frac{1}{3}$
So we know the two roots are 3.5 and $-\frac{1}{3}$.
So the equation can be written as $6x^2−19x−7 = (x-3.5)(x+1/3) = (2x−7)(3x+1)$
Ok, I think they have actually used the AC method, that's the one that makes more sense here.
For a general explanation of such method:
http://www.virtualnerd.com/algebra-1/polynomials-and-factoring/trinomial-factorization/leading-coefficient-not-1/factor-by-a-c-method
So, for this case we have: $$6x^2-19x-7$$
Hence: $$a=6, b=-19, c=-7$$
So: $$ac = -42$$
Then, we calculate the factors of 42 and their sum following the AC method (I tried to use a Latex table but it is a pain in the ass, so sorry about the format.):
$$1\times-42; 1+(-42)=41$$ $$2\times-21; 2+(-21)=-19$$
As $b=-19$ we can stop right there. Our factors are 2 and -21, then $b=2x-21x$
Now, rewriting the trinome with such factors for b, we have then:
$$6x^2+2x-21x-7$$
Now we factor by grouping by pairs: $$2x(3x+1)-7(3x+1)$$
And now, we have a $(3x+1)$ common factor there, so: $$(3x+1)(2x-7)$$
Tadaa!
PS. It is worth to add that the book itself ask explicitly to do the factorization dividing by the GCF –hence the question title–, but that's clearly impossible in this case, so it is probably an errata.