5x^2 + 11 x + blank?
5x^2 + 11 x + blank?
Hi, Leanna.
In my Algebra class we are doing something similar to this. Without the directions, it's hard to say if it's the same as our lesson.
Anyway, our book asks us to come up with possible number(s) to put in the blank so that you can factor the polynomial, 5x^2 + 11 x + blank. What we did in our math class is to write out the two sets of ( )'s and try to figure out possibilities...
(5x + ___ )(x + ___ )
Somehow we need to make the outers and inners combine to give 11x.
If we try 1 and 1, it will be too small:
(5x + 1)(x + 1)
This multiplies to give 5x^2 + 5x + 1x + 1. The outers and inners combine to give 6x (see, too small).
So how about 2 and 1?
(5x + 2)(x + 1)
This multiplies to give us 5x^2 + 5x + 2x + 2, which is 5x^2 + 7x + 2 ... this is not right because our middle term is 7x.
So, how about trying 1 and 2?
(5x + 1)(x + 2)
This multiplies to give us 5x^2 + 10x + 1x + 2. This one will work because 10x and 1x combine to give you an 11x. Ahah!
So, the answer to the blank in the original problem would be a positive 2
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.
Hey, Dan.
I tried your idea and found another answer. It was interesting to look at all the possible equations. You said to "figure out which values of K make 121 - 20K a perfect square." Well, we first have to recognize that the perfect square value cannot be higher than 121 or we would be dealing with imaginary numbers (probably beyond this student's Algebra course). So solving several equations:
121 - 20k = 100
121 - 20k = 81
121 - 20k = 64
121 - 20k = 49
121 - 20k = 36
121 - 20k = 25
121 - 20k = 16
121 - 20k = 9
121 - 20k = 4
121 - 20k = 1
The only two equations to yield whole numbers are:
121 - 20k = 81 and 121 - 20k = 1
A comment about your "Magic Number" notion - When I show students this cool trick, I call it the "Brute Force Method" because it forces the factors to show themselves. ;-)
You have neglected values of K that are negative. K = -16 makes 121 - 20K = 441 = 21^2. 5x^2 + 11x - 16 is factorable: (x - 1)(5x + 16). In reality, there are an infinite number of integral values of K that make 5x^2 + 11x + K factorable, but by the time I had done the math, so to speak, I figured no one was reading this question any more.
I don't know why I even came back to the question, but I'm glad I did. I think the point of the exercise was just to see if the student could come up with any value for k that would work. Oh, and duh... negative values for k... of course. Well, have a good night, sir.
Hi Leanna, I'm going to take a quess at what you are after here. It looks like you are trying to find a number that will allow you to right this expression as a square of another expression in the form of
Ax + B.
In other words, we want
5x^2 + 11 + Blank = (Ax + B)^2
Let's carry out the square on the right hand side:
5x^2 + 11 + Blank = (A^2)(x^2) + 2ABx + B^2
Now we can compare the coefficients one at a time:
5 = A^2 => A = sqrt(5)
11 = 2AB => B = 11/2A => B = 11/[2(sqrt(5)]
and
Blank = B^2 => Blank = 11^2/ 4(5)
I'm also guessing that your instructor probably wanted you to solve: 5x^2 + 10x + blank? because the answer would have turned out to be 5.
This operation has a purpose. It is the first step in a process called "Completing the Square" that will allow you to more easily solve quadratic equations.
Comments
I like your answer!
Thank you, John.