How do you simplify #(x^2 – 3x – 10)/(x^2 – 9)div(x^2 + 4x + 4)/(x+3)##?

Answer 1

First, factor your quadratic functions. Finding its roots will lead us to the factors and then it'll be way easier to calculate your result. So, let's find the roots (using Bhaskara)!

#f(x)=x^2-3x-10#
#(3+-sqrt(9-4(1)(-10)))/2# #(3+-7)/2# #x_1=5#, the same as the factor #x-5=0# #x_2=-2#, the same as the factor #x+2=0#
#f(x)=x^2-3x-10=(x-5)(x+2)#
Next: #g(x)=x^2-9#
#x^2=9# #x=+-3#, which generate the factors #(x-3)(x+3)#
Finally, #h(x)=x^2+4x+4#, which will have one root: #x=-2# and then give us the same factor twice: #(x+2)(x+2)#

Rewriting:

#(((x-5)(x+2))/((x-3)(x+3)))/(((x+2)(x+2))/(x+3))#=#((x-5)cancel((x+2))cancel((x+3)))/((x-3)cancel((x+3))cancel((x+2))(x+2))#=#(x-5)/((x-3)(x+2))#=#(x-5)/(x^2-x-6)#
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Answer 2

This explanation's going to be all math. I'm assuming you know how to factor polynomials, but if you don't, then don't be afraid to comment to ask how; I don't bite! Okay, let's start.

#(x^2–3x–10)/(x^2–9) ÷ (x^2+4x+4)/(x+3)#

We can flip the 2nd fraction and multiply it.

#(x^2–3x–10)/(x^2–9) times (x+3)/(x^2+4x+4)#

Now let's factor the trinomials into binomials. Hopefully this will let us cancel out some stuff.

#((x-5)(x+2))/(x^2-9) times (x+3)/((x+2)(x+2))#
We can also factor #x^2 - 9# into #x+3# and #x-3#.
#((x-5)(x+2))/((x-3)(x+3)) times (x+3)/((x+2)(x+2))#

Now, I can cancel out some of the factors to simplify the answer.

#(x-5)/(x-3) times 1/(x+2) = (x-5)/((x-3)(x+2))=(x-5)/(x^2 - x -6)#
This should be the final answer, unless your teacher prefers to show the factors of trinomials, in which case #(x-5)/((x-3)(x+2))# is also acceptable.
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Answer 3

To simplify the expression, we can start by factoring the numerator and denominator separately.

The numerator (x^2 – 3x – 10) can be factored as (x – 5)(x + 2). The denominator (x^2 – 9) can be factored as (x – 3)(x + 3). The second fraction in the expression, (x^2 + 4x + 4)/(x + 3), can be simplified to (x + 2).

Now, we can rewrite the expression as [(x – 5)(x + 2)]/[(x – 3)(x + 3)] * (x + 2)/(x + 3).

Next, we can cancel out the common factors in the numerator and denominator.

This simplifies the expression to (x – 5)/(x – 3).

Therefore, the simplified form of the expression is (x – 5)/(x – 3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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