##### the ixl topic name is called "factor quadratic equations with other leading coefficients"

 Mathematics Tutor: None Selected Time limit: 1 Day

Jun 3rd, 2015

We use the grouping method which consists on these steps:

1.) First, we multiply the first coefficient (the number in front of the s^2, the 2) times the independent number (the last number of the equation, the 11). So (2)(11) = 22.

2.) Then we need to find two numbers such as their product give 22 (the obtaine number in step 1) and their algebraic addition give 13 (the coefficient of the s, the number in front of the s). For that, we test by using the divisors of 22 (multiples of 22) which are: 1, 2, 11, and 22 (positives and negatives).

By testing some combinations, the numbers would be 2 and 11 since (2)(11) = 22 and 2 + 11 = 13.

3.) Then we use the obtained numbers in step 2 (the 2 and 11) to rewrite the middle term of the equation (the 13s).

So we have that 13s can be written like 2s + 11s. Then our expression looks like this:

2s^2 + 13s + 11 = 2s^2 + 2s + 11s + 11

4.) After this, we find the greatest common factor of 2s^2 and 2s. We always choose the s with the less exponent, so it would be s. By the way, we have another common factor which is the 2 since we have the 2 twice in the expression 2s^2 + 2s. So the greatest common factor is 2s.

On the other hand, the last two terms which are 11s + 11 we can see that we have a common factor which is 11

since we have the 11 twice.

5.) So we factor 2s out from the expression 2s^2 + 2s which would be:  2s(s + 1). On the other hand, we factor 11 out from the expression 11s + 11 which is: 11(s + 1). So now the whole expression would look like this:

2s^2 + 2s + 11s + 11 = 2s(s + 1) + 11(s + 1)

6.) From the last expression we can see a common factor which is (s + 1) since we have it twice:

2s(s + 1) + 11(s + 1). So then we factor the (s + 1) from the whole expression like this:

2s(s + 1) + 11(s + 1) = (2s + 11)(s + 1)

Jun 3rd, 2015

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Jun 3rd, 2015
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Jun 3rd, 2015
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