c What solutions make the left-hand side of the equation equal?
D
d Rewrite both solutions as fractions. What can you say about the product of the solution's numerators and denominators?
A
a (5x+3) and (x-2)
B
b x_1=- 3/5, x_2=2
C
c See solution.
D
d See solution.
Practice makes perfect
a To factor this expression we will use a generic rectangle. We know that 5x^2 and - 6 goes into the lower left and upper right corner of the generic rectangle.
To fill in the remaining two rectangles we need two x-terms that have a sum of - 7x and a product of - 30x^2.
Notice that the product is negative, which means one factor must be negative and the other must be positive. With this in mind, we can factor - 30x^2 in as many ways as we can and then add the factors. When we have found factors that add to - 7x, we have the correct ones.
c|c|c|r|c
Product & ax(bx) & ax+bx& Sum & Equals- 7x? [0.2em]
[-1em]
- 30x^2 & - x(30x) & - x+30x& 29x & * [0.1em]
- 30x^2 & - 30x(x) & -30x+x& - 29x & * [0.1em]
- 30x^2 & - 2x(15x) & - 2x+15x& 13x & * [0.1em]
- 30x^2 & - 15x(2x) & - 15x+2x& - 13x & * [0.1em]
- 30x^2 & - 3x(10x) & - 3x+10x& 7x & * [0.1em]
- 30x^2 & - 10x(3x) & -10x+3x& - 7x & âś“
When one term is 3x and the other is - 10x, we have a product of - 30x^2 and a sum of - 7x. Now we can complete the diamond and generic rectangle.
To factor the quadratic expression we add each side of the generic rectangle and multiply their sums.
5x^2-7x-6=(5x+3)(x+(-2))
⇓
5x^2-7x-6=(5x+3)(x-2)
c When we solved the equation in Part B, we set the factors equal to zero and solved for x. These solutions are the values of x that make the original equation equal to zero.
d The leading coefficient is 5 and the constant is -6. If we rewrite both solutions as fractions, we see that the leading coefficient is the product of the denominators and the constant is the product of the numerators.
