SOLUTION: can u solve this problem factorise 15+15x-20x^2

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Question 221303: can u solve this problem
factorise
15+15x-20x^2
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Can you solve this problem

Factorise

15%2B15x-20x%5E2

Step 1. Factor out -5 and rearrange in descending order

15%2B15x-20x%5E2=+-5%284x%5E2-3x-3%29 ANSWER

Step 2. Now work with the terms in the parenthesis. We need two integers m and n whose sum is -3 and whose product m*n=(-3)(4)=-12.

Step 3. After trial and error there are no integers that satisfies the quadratic equation.

Step 4. Use the quadratic formula given below for the equation 4x%5E2-3x-3

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=4, b=-3, and c=-3

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B-3x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-3%29%5E2-4%2A4%2A-3=57.

Discriminant d=57 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+57+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+57+%29%29%2F2%5C4+=+1.31872930440884
x%5B2%5D+=+%28-%28-3%29-sqrt%28+57+%29%29%2F2%5C4+=+-0.568729304408844

Quadratic expression 4x%5E2%2B-3x%2B-3 can be factored:
4x%5E2%2B-3x%2B-3+=+4%28x-1.31872930440884%29%2A%28x--0.568729304408844%29
Again, the answer is: 1.31872930440884, -0.568729304408844. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-3%2Ax%2B-3+%29




Step 5. ANSWER: 15%2B15x-20x%5E2=+-5%284x%5E2-3x-3%29 and we could use the quadratic equation to factor out the quadratic terms as found in step 4.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J