McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 63 Page 83

Start by identifying the values of a, b, and c. Be sure that all of the terms are on the same side and in the correct order for the standard form of a quadratic function.

- 1/2 and 5/3

Practice makes perfect
To solve the given equation by factoring, we will start by identifying the values of a, b, and c. 6x^2-7x-5=0 ⇔ 6x^2+( - 7)x+( - 5)=0 We have a quadratic equation with a= 6, b= - 7, and c= - 5. To factor the left-hand side we need to find a factor pair of 6 * - 5=- 30 whose sum is - 7. Since - 30 is a negative number, we will only consider factors with opposite signs — one positive and one negative — so that their product is negative.
Factor Pair Product of Factors Sum of Factors
1 and - 30 ^(1* (- 30)) - 30 1+(- 30) - 29
- 1 and 30 ^(- 1* 30) - 30 - 1+30 29
2 and - 15 ^(2* (- 15)) - 30 2+(- 15) - 13
- 2 and 15 ^(- 2* 15) - 30 - 2+15 13
3 and - 10 ^(3* (- 10)) - 30 3+(- 10) - 7
- 3 and 10 ^(- 3* 10) - 30 - 3+10 7
The integers whose product is - 30 and whose sum is - 7 are 3 and - 10. With this information, we can rewrite the linear factor on the left-hand side of the equation, and factor by grouping.
6x^2-7x-5=0
6x^2+3x-10x-5=0
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Factor out 3x & - 5
3x(2x+1)-10x-5=0
3x(2x+1)-5(2x+1)=0
(3x-5)(2x+1)=0
Now we are ready to use the Zero Product Property.
(3x-5)(2x+1)=0
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Solve using the Zero Product Property
lc3x-5=0 & (I) 2x+1=0 & (II)
lc3x=5 & (I) 2x+1=0 & (II)
lcx = 53 & (I) 2x+1=0 & (II)
lcx = 53 & (I) 2x=- 1 & (II)
lcx = 53 & (I) x= - 12 & (II)
lcx = 53 & (I) x=- 12 & (II)
We found that the solutions to the given equation are x= 53 and x=- 12. To check our answer, we will graph the related function y=6x^2-7x-5 using a calculator.

We can see that the x-intercepts are - 0.5, or - 12, and 53. Therefore, our solutions are correct. âś“