# Simplify the algebraic fraction $\frac{4x^2-8x+3}{4x^2+2x-12}$.

Find the quotients as algebraic fractions of the following division:

$$\frac{4x^2-8x+3}{4x^2+2x-12}$$

$$\frac{2x-1}{2(x+2)}$$

The method that I am applying to solve:

Write what $$2$$ numbers multiplied give me the far most right number in the numerator, and add, give me the number left to the far most right number in the numerator as the numerator for the new fraction and doing the same for the denominator then cancelling out whichever $$2$$ numbers i can.

Ex.

1. $$\frac{x^2 -2x -3}{x^2 -1}$$
2. $$\frac{(x+1)(x-3)}{(x+1)(x-1)}$$
3. after I cancel out $$(x+1)$$, I’m left with $$\frac{x-3}{x-1}$$

I can't find any numbers that both multiply together to get $$3$$ and add together to get $$-8$$. i tried simplifying the original fraction then applying the method but i still didn't get the right answer. This method should work because it worked for part a) and b) in the question

• You seem to be ignoring the "$4$" in "$4x^2$". May 7, 2020 at 2:58
• What you need to do is factor the numerator and denominator and cancel any common factors. $$4x^2-8x+3=(2x-1)(2x-3)$$ and $$4x^2+2x-12=2(2x^2+x-6)=(2x-3)(x+2)\;,$$ so we have $\frac{(2x-1)(2x-3)}{2x-3)(x+2)}$ and can cancel the common factor of $2x-3$ to leave $\frac{2x-1}{2(x+2)}$. May 7, 2020 at 2:59

$$4x^2 - 8x + 3$$

When you have a coefficent on the $$x^2$$ term, it gets a little bit more complicated to factor. There are two main techniques. Since 3 is prime, we can start with the guess

$$(Ax - 3)(Bx - 1)$$

Now we look for $$A,B$$ such that $$AB = 4$$ and $$-A - 3B = -8$$

Our candidates are $$\{(1,4),(2,2), (4,1)\}$$ and we try these until something hits.

The other way students are taught, is more generalizable, but in this case more work.

$$4x^2 - 8x + 3$$

We take the first coefficent and the last and multiply them together.

$$4\cdot 3 = 12$$

Now we look for two numbers that when added together make $$-8$$ and multiplied together make $$12.$$ This is exactly as you describe above.

$$(-2),(-6)$$ work.

Now we say:

$$4x^2 - 2x - 6x + 3$$

group the terms:

$$(4x^2 - 2x) + (-6x + 3)$$

The expression on the left has a common factor of $$2x$$ and the expression on the right has a common factor of $$3.$$

$$2x (2x - 1) + 3(-2x + 1)$$

Let's factor our a $$-1$$ from the expression on the right to flip some signs and make the common factor between the two stand out.

$$2x (2x - 1) - 3(2x - 1)$$

And factor out the common factor

$$(2x-3)(2x - 1)$$

In the denominator, before you start, notice that $$2$$ is a common factor to all three terms.

$$4x^2 +2x - 12 = 2(2x^2 + x - 6)$$ and then proceed as above.

We might guess that $$(2x - 1)$$ or $$(2x - 3)$$ will be a factor, in order to have something that canceles. While this isn't "mathematically rigorous" logic, thinking like a teacher, test designer or textbook publisher often points to short-cuts.

Once you have the answer you can just divide the numerators to find what was divided out, so $$\frac{4x^2-8x+3}{2x-1}=2x-3$$ In your try you seem to have been ignoring the factor $$4$$ on the $$x^2$$ term.