Question 27737: Factor:
9x^2-12x-5
How do I do this ?
Thanks for your help and time !
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! Factor:
9x^2-12x-5
To factorise 9x^2-12x-5 means to express this given quadratic expression as a product of two linear factors.
The Rule: Multiply the coefficient of the square term and the constant term along with the signs, take the numerical product, find its factors, group them into two sets in such a way that their sum is the mid term coefficient (midterm is the term in x)(while grouping the factors to form the two required sets, care should be taken to include every factor: that is, you should not omit a single factor).Then
if the sign of the product is minus, and the middle term (the term in x) if minus,then give the sign of the midterm(the term in x) to the larger set and the other sign to the smaller
Now in the given expression (9x^2-12x-5)
The product of the coefficient of x^2 and the constant term (-5)
is (9)X(-5) = -45 = -(1X3X3X5)
The numerical factors are 1,3,3 and 5
You may leave out the 1 as (1X anything) =the same thing
Grouping the factors into two sets without omitting any factor, the sets are
(3X5) and 3 so that You observe that (-15)+(3) = -12 and hence write the midterm -12x = -15x+3x.
Therefore (9x^2 -12x -5)
= 9x^2 +(-15x +3x)-5
= 9x^2 -15x +3x -5 (by additive associativity)
= 3x(3x-5)+1(3x -5)(by taking 3x common in the first two terms and 1 in the other two)
=(3x)p +p where p = (3x-5)
=p(3x+1)
= (3x-5)(3x+1)
Answer: (9x^2 -12x -5)= (3x-5)(3x+1)
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