SOLUTION: 9p^2+6p-8

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Question 1210401: 9p^2+6p-8

Found 5 solutions by ikleyn, josgarithmetic, mccravyedwin, greenestamps, Edwin McCravy:
Answer by ikleyn(52834)   (Show Source): You can put this solution on YOUR website!
.
9p^2+6p-8
~~~~~~~~~~~

If you want to get an answer,  you should first ask a question,
in a proper, accurate, clear, unambiguous and polite way.



Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!
A few combinations and tries give .
If not what you are looking for then you need to specify.
Answer by mccravyedwin(408)   (Show Source): You can put this solution on YOUR website!
In case the instructions were "factor" or "factorise" the quadratic.
the answer is in the form

(Ap+B)(Cp-D)

Since BD must be negative, we choose + for the first parentheses and - for 
the second, so that when we FOIL, the last term will be "-".

We choose positive integers A, B, C, and D, so that 
AC = 9 and BD = 8. 

(p+1)(9p-8) = 9p²+p-8
(p+2)(9p-4) = 9p²+14p-8
(p+4)(9p-2) = 9p²+34p-8
(p+8)(9p-1) = 9p²+71p-8
(3p+1)(3p-8) = 9p²-21p-8
(3p+2)(3p-4) = 9p²-6p-8
(3p+4)(3p-2) = 9p²+6p-8
(3p+8)(3p-1) = 9p²+21p-8
(9p+1)(p-8) = 9p²-71p-8
(9p+2)(p-4) = 9p²-34p-8
(9p+4)(p-2) = 9p²-14p-8
(9p+8)(p-1) = 9p²-p-8

Only one of them is equal to the given quadratic.  
Can you find which one it is?

Edwin
Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


Here is one of many ways to factor a quadratic like this, where the leading coefficient is not 1.

Step 1: find the product of the leading coefficient and the constant term
9(-8) = -72

Step 2: find two integers (obviously one positive and one negative) whose product is -72 (from step 1) and whose sum is 6 (the coefficient of the linear term)
-72 = (12)(-6)

Step 3: break the linear term into two terms using the numbers from step 2
9p^2+12p-6p-8

Factor the new expression by grouping
(9p^2+12p)+(-6p-8)
3p(3p+4)+(-2)(3p+4)
(3p-2)(3p+4)

ANSWER: 9p^2+6p-8 = (3p-2)(3p+4)
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

It's just about as easy to solve the quadratic equation









, 

, 

Then clear the fractions by multiplying each parentheses by 3,

, 



Edwin
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