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by sureshbala » Tue May 05, 2009 10:49 pm
The given question must be this........

If y is not equal to -7, then find the value of

(y^3 + 5y^2 - 15y - 7)/(y + 7)

= (y + 7)(y^2 - 2y - 1)/(y + 7)

= y^2 - 2y - 1
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by Vemuri » Tue May 05, 2009 11:03 pm
sureshbala wrote:
(y^3 + 5y^2 - 15y - 7)/(y + 7)

= (y + 7)(y^2 - 2y - 1)/(y + 7)

= y^2 - 2y - 1
Hi Suresh, can you please clarify how you derived (y + 7)(y^2 - 2y - 1) from (y^3 + 5y^2 - 15y - 7)?
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by sureshbala » Tue May 05, 2009 11:16 pm
Vemuri wrote:
sureshbala wrote:
(y^3 + 5y^2 - 15y - 7)/(y + 7)

= (y + 7)(y^2 - 2y - 1)/(y + 7)

= y^2 - 2y - 1
Hi Suresh, can you please clarify how you derived (y + 7)(y^2 - 2y - 1) from (y^3 + 5y^2 - 15y - 7)?
Hi, this is all anticipation.........

Since there is y+7 in the denominator, I verified whether the numerator is a multiple of y+7 are not.

So divide the given polynomial with y+7 and you will find that remainder is 0.
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by prince11 » Fri Jun 19, 2009 1:46 pm
Hi Suresh/friends,

can you explain how (y^3 + 5y^2 - 15y - 7) became (y^2 - 2y - 1).
I know that (y + 7) (y^2 - 2y - 1)/(y + 7) = (y^3 + 5y^2 - 15y - 7), but how do I go from (y^3 + 5y^2 - 15y - 7) and (y + 7) to get (y^2 - 2y - 1).

Thanks
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by cramya » Fri Jun 19, 2009 6:48 pm
can you explain how (y^3 + 5y^2 - 15y - 7) became (y^2 - 2y - 1).
I know that (y + 7) (y^2 - 2y - 1)/(y + 7) = (y^3 + 5y^2 - 15y - 7), but how do I go from (y^3 + 5y^2 - 15y - 7) and (y + 7) to get (y^2 - 2y - 1).

Use Polynomial Long Division.

This might help

https://www.purplemath.com/modules/polydiv2.htm

Let us know if u still hv questions.

Regards,
CR
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by anksgupta » Fri Jun 19, 2009 6:59 pm
A still quicker approach is to use synthetic division for polynomials

https://www.purplemath.com/modules/synthdiv.htm
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by prince11 » Sun Jun 21, 2009 8:46 am
Thank you cramya and anksgupta.
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by Stuart@KaplanGMAT » Sun Jun 21, 2009 9:19 am
sureshbala wrote:The given question must be this........

If y is not equal to -7, then find the value of

(y^3 + 5y^2 - 15y - 7)/(y + 7)

= (y + 7)(y^2 - 2y - 1)/(y + 7)

= y^2 - 2y - 1
The key to finding this solution (which is great) is always checking out the answer choices and understanding how the GMAT works.

You do not need to know how to factor complex polynomials for the GMAT. If you see a question that involves complex polynomials, there will always be a way to solve it without knowing wild and wacky formulas.

In this case, we certainly could have solved the problem quickly and easily by picking numbers.

Alternatively, we could look at the answer choices and see (at least I assume this is what we'd see, since the OP didn't include the choices - boo!) that none of them are in fraction form. In other words, the denominator in the original expression disappears.

Since the fraction disappears, we know that (y + 7) is going to cancel out. We proceed to factor based on this knowledge.

Now when we factor out:

(y^3 + 5y^2 - 15y - 7)

we know that it will be in the form:

(y + 7)(something)

and once we have that first bracket the rest of the exercise is simple.

We see that the final term is -7, so the final term of the second bracket needs to be -1.

We see that the first term is y^3, so the first term of the second bracket needs to be y^2.

So, we currently have:

(y + 7)(y^2 +ky -1)

and now we need to pick a value for k that will give us both 5(y^2) and -15y.

We already have 7(y^2) (from 7*y^2), so y(ky) must equal -2(y^2). Therefore, k=-2.

Accordingly, the entire expression is:

(y + 7)(y^2 - 2y - 1)/(y + 7)

which of course simplifies to:

(y^2 - 2y - 1)

as shown by sureshbala.

Note that we just used some common sense, logic and basic multiplication to solve.
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by wishkaro » Mon Jun 22, 2009 1:20 am
Stuart Kovinsky Great explaination !!

Thanks :)
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