^{2}+18X + 5

^{2 }+15X + 3X +5 /Break up 18X = 15X +3X

^{ }3X + 5 = 0 X = -5/3

^{2 }+ bX +c

I cannot figure out how to get x from this problem. Thanks a lot!

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

f(x) =9X^{2} +18X + 5

First check factorability of the quadratic, find out whether there exist 2 integers , a, b, such that

their Sum equals 18, and their product is 9*5 =45

To do that :

Write 45 as product of its factors: 45 = 3 * 15 = 9 * 5

Observe that 3+ 15 =18

f(X) = 9X^{2 }+15X + 3X +5 /Break up 18X = 15X +3X

3X(3X + 5 ) +(3x +5) / factor the quadratic by grouping

( 3X + 5 )( 3X + 1) =0

3X + 1 = 0 X = -1/3

In the event if, in quadratic aX^{2 }+ bX +c

there is no 2 integers , m;n, does not exist such that m.n= ac and m+n = b

Then to solve the quadratic you can either use factoring by completing square or

using quadratic formula.

Hi William;

9x^{2} + 18x + 5 =0

For the FOIL...

FIRST must be (3x)(3x) or (9x)(x).

OUTER and INNER must add-up to 18x.

LAST must be (5)(1) or (1)(5).

(3x+5)(3x+1)=0

FOIL...

FIRST...(3x)(3x)=9x^{2}

OUTER...(3x)(1)=3x

INNER...(5)(3x)=15x

LAST...(5)(1)=

9x^{2}+3x+15x+5=0

9x^{2}+18x+5=0

So...

(3x+5)(3x+1)=0

One or both of the parenthetical equations must equal zero.

3x+5=0

3x=-5

x=-5/3

and/or

3x+1=0

3x=-1

x=-1/3

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