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- Jul 18, 2013

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- Thread starter Rido12
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You can directly follow the hint

\(\displaystyle \lim_{x \to -2}3x^2+ax+a+3=0\)

\(\displaystyle \lim_{x \to -2}3x^2+ax+a+3=0\)

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\(\displaystyle \lim_{x \to 2}\frac{x^2-a}{x-2}\)

So , what should be the value of $a$ so that the limit exists and why ?

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(x-2), right?

- Aug 18, 2013

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Hint #1: It is easy to see that the denominator factors as $(x-1)(x+2)$. Since the limit goes to $x=-2$, we must get rid of the $(x+2)$ term in the denominator. To do this, the numerator also has to factor with an $(x+2)$ term.

Hint #2: The leading coefficient of the numerator is 3, and one of its factors ($(x+2)$) has a leading coefficient of 1, so the other factor has a leading coefficient of 3, so it is in the form $(3x+k)$, for some constant k.

Hint #4: Expanding $(x+2)(3x+k)$ gives $3x^2+(6+k)x+2k$, which in turn gives $6+k=a$ and $2k=a+3$. Using this, it is easy to solve for a and k, which will give the factors of the numerator.

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Actually, I'm pretty sure he meant a constant term, so no. Try to find an $a$ such that the numerator contains an $(x-2)$ term.(x-2), right?

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The problem lies on Hint two; I lost you there D:Hint #2: The leading coefficient of the numerator is 3, and one of its factors ($(x+2)$) has a leading coefficient of 1, so the other factor has a leading coefficient of 3, so it is in the form $(3x+k)$, for some constant k.

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Actually, I'm pretty sure he meant a constant term, so no. Try to find an $a$ such that the numerator contains an $(x-2)$ term.

Then a could be 4?Actually, I'm pretty sure he meant a constant term, so no. Try to find an $a$ such that the numerator contains an $(x-2)$ term.

Last edited:

- Aug 18, 2013

- 76

We want the numerator to factor as $3x^2+ax+a+3=(x+2)(mx+k)$ for some coefficients m and k. Expanding this gives $3x^2+ax+a+3=mx^2+kx+2mx+2k$. Therefore, $m=3$ and restrictions to $a$ can be made.The problem lies on Hint two; I lost you there D:

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Yes, it is. The numerator can factor as \(\displaystyle (x+2)(x-2)\)Then a could be 4?

*Edit* Actually, you don't even need to find

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Thanks, I get it now, but, what led you choose "(mx+k)"? Could you not just have done (x+2)(x+k)?We want the numerator to factor as $3x^2+ax+a+3=(x+2)(mx+k)$ for some coefficients m and k. Expanding this gives $3x^2+ax+a+3=mx^2+kx+2mx+2k$. Therefore, $m=3$ and restrictions to $a$ can be made.

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Yes, it is. The numerator can factor as \(\displaystyle (x+2)(x-2)\)

*Edit* Actually, you don't even need to findrestrictionsto a. You can solve for it directly using Hint #4.

EDIT: Oh, could "m" be representing any arbitrary number?

- Aug 18, 2013

- 76

Expanding $(x+2)(x+k)$ gives $x^2+(k+2)x+2k$, where the polynomial we want is actually $3x^2+ax+a+3$. Note that the coefficient of $x^2$ in the cases are different. If we want the equality to be satisfied forThanks, I get it now, but, what led you choose "(mx+k)"? Could you not just have done (x+2)(x+k)?

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It is a variable which we solved for.Oh, could "m" be representing any arbitrary number?

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- #11

- Jul 18, 2013

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If I wanted it in the form $3x^2+ax+a+3$, and I had to multiple (something)(x+2) to obtain it in that form, it wasn't intuitive for me to multiple that by "(mx+K)". It works out, but I wouldn't have came up with that immediately ... so would I have to use like trial and error to do so, or is there some sort of principle?Expanding $(x+2)(x+k)$ gives $x^2+(k+2)x+2k$, where the polynomial we want is actually $3x^2+ax+a+3$. Note that the coefficient of $x^2$ in the cases are different. If we want the equality to be satisfied forallx, we want the coefficients to be the same.

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It is a variable which we solved for.

- Aug 18, 2013

- 76

What's wrong with \(\displaystyle (mx+k)\)?It works out, but I wouldn't have came up with that immediately ... so would I have to use like trial and error to do so, or is there some sort of principle?

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- #13

- Jul 18, 2013

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Nothing, it's just that I wouldn't have come up with that straight off the bat. If I needed it in the form 3x2+ax+a+3, multiplying (x-2) by "mx+k" wouldn't have been apparent. How were you able to determine so quickly that "mx+k" was the ideal candidate?What's wrong with \(\displaystyle (mx+k)\)?

- Aug 18, 2013

- 76

The polynomial you were trying to factor was of degree 2, so there are two factors of degree 1, that is, of the form $(ax+b)(cx+d)$.How were you able to determine so quickly that "mx+k" was the ideal candidate?

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- #15

- Jul 18, 2013

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Thank you!The polynomial you were trying to factor was of degree 2, so there are two factors of degree 1, that is, of the form $(ax+b)(cx+d)$.