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## Licensed HS Math Teacher, 13 years experience

### Valparaiso, IN (46385)

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# Dan's Responses in WyzAnt Answers

#### factoring 5x^2 +11x + blank?

5x^2 + 11 x + blank?

+ more- less
Asked by Leanna from Lorain, OH
10

I disagree with John.  It sounds like the question is asking: For which values of K is the quadratic trinomial p(x) = 5x^2 + 11x + K factorable"?

Let us assume that p(x) = 5x^2 + 11x + K is factorable.  Since 5 and 11 have no common factor other than 1, we know that we cannot factor out a Greatest Common Factor.  Then p(x) = (ax + b)(cx + d) for 4 real numbers a, b, c, and d.  And since we do not normally factor unless a, b, c, and d can be written as integers, let us also assume that a, b, c, and d are integers, possibly negative.

If we simulate polynomial multiplication on (ax + b)(cx + d) we get:

5x^2 + 11x + K = (ac)x^2 + (ad + bc)x + (bd)

Equalizing coefficients we get:

(ac) = 5, (ad + bc) = 11, and (bd) = K

Since 5 is prime and we are assuming that a, b, c, and d are all integers, either a = 1 and c = 5 or a = 5 and c = 1.  It does not matter which, so let us assume that a = 1 and c = 5.  This leaves us with:

5x^2 + 11x + K = (1x + b)(5x + d)

To complete the factorization of p(x); factor by grouping.  I always teach my students to use the "magic number" technique.  Multiply the x^2 term and the constant term.  The reason I do this is because the coefficient of the x^2 term is (ac) and the constant term is (bd).  So their product is (abcd)x^2, the coefficient of which contains all 4 values a, b, c, and d.  Notice also that the coefficient of the x term also contains all 4 values a, b, c, and d.  So now factoring p(x) amounts to finding factors of the "magic number" whose sum is the coefficient of the middle term.  This allows you to split the a, b, c, and d into the correct pairings.

I.e. (abcd) = 5K and (ad + bc) = 11

For p(x) = 5x^2 + 11x + K, this process yields a magic number of 5K and a sum of 11.  So go through the multiples of 5 and every time you get a multiple of 5 which has a pair of factors that adds up to 11, that corresponding value of K is one of the values that makes p(x) factorable.

For example, if K = 1, the magic number is 5(1) = 5.  5 does not have a pair of factors whose sum is 11, so K = 1 is not one of the solutions.  If K = 2, the magic number is 5(2) = 10.  10 does have a pair of factors whose sum is 11, 10 and 1.  So K = 2 is one of the solutions.  There are 2 more solutions for K that I found.  Let me know when you have found the other 2.

A second way to solve this problem is the consider the quadratic equation 5x^2 + 11x + K = 0.  The polynomial p(x) = 5x^2 + 11x + K is factorable only if the solutions to 5x^2 + 11x + K = 0 are rational numbers.  Using the quadratic formula;

a = 5, b = 11, c = K yields the following solutions:

x = (-11 + sqrt(121 - 20K))/10 and x = (-11 - sqrt(121 - 20K))/10.  The solutions to this equation will be rational only if the value of 121 - 20K is a perfect square.  So figure out which values of K make 121 - 20K a perfect square.  Again, there are 3 solutions.  The same three you get using the other method I suggested.

Not thinking.  There is a 4th solution when K = 0.

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