Question

Okay can someone figure out how to factor and cancel this equation. It's a limits prob but my main question is I been trying to figure out how they factor the equation out (common factor) and what had been canceled.

Answer 1

maybe start by factoring numerator and denominator

Answer 2

@Jhannybean

Answer 3

@mathstudent55

Answer 4

I'm learning these kinds of problems too. Let me attempt to see what they did.. the arrow points to the answer, correct.?

Answer 5

Let's start by writing out the function again.. \[\frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}\]Let's start by the numerator. Factor out the LCM first, in this case, our lowest common factor is \(\color{red}{x^2}\).\[x^2(x^2+5x+6)\]Now what are two numbers that multiply to give 6 and add to give 5? that would be \(\color{blue}{3~,~2}\) Let's now put the numerator into full factor form. \[x^2(x+2)(x+3)\]Now let's work on our denominator: \(x^2\color{red}{(x+1)}-4\color{red}{(x+1)}\)

Answer 6

Notice how there is a common base \((x+1)\) here highlighted in red? we can factor that out, which leaves us with: \((x^2-4)(x+1)\)

Let's rewrite out entire function combining everything we have simplified.\[\frac{x^2(x+3)(x+2)}{\color{red}{(x^2-4)}(x+1)}\]\(x^2-4\) can be reduced as well. By identifying it as a perfect square, we can say \((x)^2 -(2)^2 = (x-2)(x+2)\) as it is the difference of two squares. Rewrite your function again. \[\frac{x^2(x+3)\color{red}{(x+2)}}{\color{red}{(x+2)}(x-2)(x+1)}\] The portions in red can cancel out, leaving you with: \[\frac{x^2(x+3)}{(x-2)(x+1)}\]

Answer 7

Now expanding both numerator and denominator, you will find that you get: \[\frac{x^3+3x^2}{x^2-x-2}\]

Answer 8

WOW^

Your really good at that