I would like find out how to get the middle and last term together

I would like find out how to get the middle and last term together

Tutors, sign in to answer this question.

To factor 40 z^2 - 47 z + 12 set up two sets of parentheses with 40z as the first term in both cases. Each of these grouips will have a minus sign following the 40z term since the product must be positive and the the integers add up to -47. At the same time, draw a horizontal line underneath both of these binomials and place a 40 as the denominator. Eventually we may need to factor the 40 into 2 * 20 or 4 * 10 or 5 * 8,. but we do not know in advance what it will be.

Now, we need to find two integers (both of which are negative) whose sum is -47 while their product is 40 * 12 or 480.

Since 40 + 12 makes 52, try 20 and 24 (take half of the larger and twice of the smaller to obtain a pair of integers that will certainly have the same product). Since 20 + 24 is only 44 we need integers that are further apart (but not as far apart as 40 and 12). I multiplied 20 by 3/2 and 24 by 2/3 to obtain 30 and 16 (whose sum is 46). Next, I doubled 16 and took half of 30 to get 32 and 15. I inserted the 32 into one group and 15 into the other. (It does not matter where since both groups are the same (40z - ) and (40z - ). Finally 40z - 32 can be factored by 8 while 40z - 15 can be factored by 5. So, we now change the 40 in the denominator to 8 * 5 and simplify to (5z -4) (8z - 3).

Setting each of these binomials to zero, we obtain z = 4/5 or z = 3/8

One does not need to use the quadratic formula, but that would certainly be another method to arrive at the same result.

There are several ways of solving quadratics: the quadratic formula, completing the square, square root, factoring. If your lesson is asking you to use a specific method you should use that one. In this case, I think you should just use the quadratic formula since you will not be able to factor out 47z.

So, first recall the equation is in the form ax^{2}+bx+c=0 and the quadratic formula is x=-b±√b^{2}-4ac / 2a. There are two solutions since there is a ± in the formula.

So, plug in the numbers from your situation: x=-(-47)±√(-47)^{2}-4(40)(12) / 2(40)

Now work out the numbers: x=47±√2209-1920 / 80

Take it further: x=47±√289 / 80

Once more: x=47±17 / 80

Finally: x=64/80 or x=30/80 which are reducible to x=4/5 or x=3/8

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.