1. ## cannot factor y^2-3y-4

I have to solve the following equation.

$\frac{3}{y-4}-\frac{2}{y+1}=\frac{5}{y^2-3y-4}$

I suspect the LCD (lowest common denominator) is $(y-4)(y+1)$ and some factor of $y^2-3y-4$ But I couldn't factor this last expression.

According to the book answers the solution is -6

2. Originally Posted by Alienis Back
I have to solve the following equation.

$\frac{3}{y-4}-\frac{2}{y+1}=\frac{5}{y^2-3y-4}$

I suspect the LCD (lowest common denominator) is $(y-4)(y+1)$ and some factor of $y^2-3y-4$ But I couldn't factor this last expression.

According to the book answers the solution is -6
$y^2-3y-4$ does factorise because the discriminant is a perfect square.

It factorises into (y-4)(y+1) which makes your 'question':

$\frac{3}{y-4}-\frac{2}{y+1}=\frac{5}{(y-4)(y+1)}$

Therefore the LCD is (y-4)(y+1). To go on and solve it:

$\frac{3(y+1) - 2(y-4)}{(y-4)(y+1)} = \frac{5}{(y-4)(y+1)}$

The denominator will cancel so then all one needs to do is expand and solve

3. Originally Posted by Alienis Back
I have to solve the following equation.

$\frac{3}{y-4}-\frac{2}{y+1}=\frac{5}{y^2-3y-4}$

I suspect the LCD (lowest common denominator) is $(y-4)(y+1)$ and some factor of $y^2-3y-4$ But I couldn't factor this last expression.

According to the book answers the solution is -6
???? (y-4)(y+1) is a factorization of $y^2- 3y- 4$!

4. Wow...wait. How did you do that? I mean the formulae for the perfect square trinomials are

$(a+b)^2=a^2+2ab+b^2$ or

$(a-b)^2=a^2-2ab+b^2$

Which is the same as:
$a^2+2ab+b^2=(a+b)^2$ or

$a^2-2ab+b^2=(a-b)^2$

And then I thought of $y^2-3y-4$ as being equivalent
to $y^2-3y-2^2$ and therefore:

$y^2-3y-4$ should have been equal to $(y-2)^2$ but this didn't fit into my solution.

After that I noticed that according to the formulae, in a perfect square, the last term, b, should alway be positive while mine, -4, is negative. How did you know it is a perfect square trinomial when it doesn't match these formulae and how did you factor it??
Just hope there are not too many questions...It is only that an answer now will solve many questions in the future.

5. Originally Posted by Alienis Back
Wow...wait. How did you do that? I mean the formulae for the perfect square trinomials are

$(a+b)^2=a^2+2ab+b^2$ or

$(a-b)^2=a^2-2ab+b^2$

Which is the same as:
$a^2+2ab+b^2=(a+b)^2$ or

$a^2-2ab+b^2=(a-b)^2$

And then I thought of $y^2-3y-4$ as being equivalent
to $y^2-3y-2^2$ and therefore:

$y^2-3y-4$ should have been equal to $(y-2)^2$ but this didn't fit into my solution.

After that I noticed that according to the formulae, in a perfect square, the last term, b, should alway be positive while mine, -4, is negative. How did you know it is a perfect square trinomial when it doesn't match these formulae and how did you factor it??
Just hope there are not too many questions...It is only that an answer now will solve many questions in the future.
I'm not entirely sure what you're using but you can find out if any given quadratic has two equal roots if the discriminant is equal to 0. That is for $f(x) = ax^2+bx+c=0$ then f(x) will be a perfect square if $b^2-4ac=0$

6. Originally Posted by Alienis Back
And then I thought of $y^2-3y-4$ as being equivalent
to $y^2-3y-2^2$ and therefore:

$y^2-3y-4$ should have been equal to $(y-2)^2$ but this didn't fit into my solution.

After that I noticed that according to the formulae, in a perfect square, the last term, b, should alway be positive while mine, -4, is negative. How did you know it is a perfect square trinomial when it doesn't match these formulae and how did you factor it??
I'm not sure why you thought that $y^2-3y-4$ is a perfect square trinomial. As an example, $y^2-4y+4$ is a perfect square trinomial because
(1) $y^2$ is a perfect square,
(2) The constant term, 4 is perfect square (and it must be positive), and
(3) the middle term, 4y, is twice the product of y and 2, the square roots of the two outer terms. (This can be negative.)
So $y^2-4y+4 = (y - 2)^2$.

Since $y^2-3y-4$ is not a perfect square trinomial, you have to find the factors by guess-and-check.

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