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## 2x^2+23x+11

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# 2 Answers

Problems such as this (factoring quadratic expressions) can be easily solved by a technique called grouping. First factor out any common constant factors out. In the given problem there are no common constant factors, so we proceed to next step.

The coefficient of x

^{2}is 2 (a in ax^2 + bx + c) and the constant is 11. First multiply 11 by 2 (a * c). 11 * 2 = 22.Now look at the coefficient for x. It is 23. Now what two factors when multiplied yield 22 and when added yield 23? 22 and 1. Now write the 23x as 22x + x. So the expression now becomes

2x^2+22x+x+11

Now group the first two terms together and the last two together and factor out common terms to make the expression look like below

2x(x + 11) + 1(x+11)

Taking the common factor x+11 out, we get the final result for the expression as

(x+11)(2x+1)

Hi Griin;

I am guessing you need this factored.

2x

^{2}+23x+11For the FOIL, we know that...

FIRST must be (2x)(x)

LAST must be (11)(1) or (1)(11)

INNER and OUTER must add-up to 23x; 2(11)x+1x=23x

So...

(2x+1)(x+11)

Let's FOIL

FIRST...(2x)(x)=2x

^{2}OUTER...(2x)(11)=22x

INNER...(1)(x)=x

LAST...(1)(11)=11

2x

^{2}+22x+x+112x

^{2}+23x+11Is this the answer you need? If not, please let me know.