Question 6428: I need help with these problems can you solve these out for me in steps?
Set 5
1. factor completely: x to the 2nd power -7x -30
2. factor completely: x to the 2nd power + 4x -16
3. factor completely: 2x to the 2nd power -11x+5
4. factor completely: 4x to the 2nd power +20x-24
5. factor completely: 4m to the 2nd power - 9
6. factor completely: 16a to the 2nd power - 25b to the 2nd power
7. solve by factoring: x to the 2nd power -x-12=0
8. solve by factoring: 2c to the 2nd power -5=-9c
9. solve the equation:(x+6)(x-7)(x-8)(x+9)=0
10. find the dimensions of the rectangle if the width is 3 feet less than the length and the area is 40ft to the 2nd power
Answer by ichudov(507) (Show Source):
You can put this solution on YOUR website! I'll do a couple for you....
Solved by pluggable solver: EXPLAIN simplification of an expression |
Your Result:
YOUR ANSWER
- This is an equation! Solutions: x=-6,x=7,x=8,x=-9.
- Graphical form: Equation
was fully solved. - Text form: (x+6)*(x-7)*(x-8)*(x+9)=0 simplifies to 0=0
- Cartoon (animation) form:
For tutors: simplify_cartoon( (x+6)*(x-7)*(x-8)*(x+9)=0 )
- If you have a website, here's a link to this solution.
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DETAILED EXPLANATION
Look at . Notes known roots (x-(-6))
Look at . Remove unneeded parentheses around factor  It becomes .
Look at . Remove extraneous '1' from product  It becomes .
Look at . Notes known roots (x-(7))
Look at . Remove unneeded parentheses around factor  It becomes .
Look at . Remove extraneous '1' from product  It becomes .
Look at . Notes known roots (x-(8))
Look at . Remove unneeded parentheses around factor  It becomes .
Look at . Remove extraneous '1' from product  It becomes .
Look at . Remove unneeded parentheses around terms , It becomes .
Look at . Solved linear equation equivalent to x+9 =0 It becomes . Result: 
This is an equation! Solutions: x=-6,x=7,x=8,x=-9.
Universal Simplifier and Solver
Done!
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Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=81 is greater than zero. That means that there are two solutions: .


Quadratic expression can be factored:

Again, the answer is: 5, 0.5.
Here's your graph:
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