## lildancngurl 3 years ago No matter how hard I try I can not figure out how to factor this: 4a^2-32a+63. Any help please? Thanks so much. Please explain the steps!

1. znimon

(2 a-9) (2 a-7)

2. znimon

So take the coefficient on the a squared term and multiply that to 63 you need to find two number (we will call them A and B) such that A * B = 4 * 63 and A + B = -32

3. znimon

you do this with other polynomials but the coefficient on the squared term is usually a 1so most people don't teach it that way.

4. lildancngurl

You are fabulous. Thank you so much! No matter how hard I tried, I could not understand it. You've made it so much simpler!

5. znimon

6. lildancngurl

I do have one more question...The two numbers I got were 14 and 18.. How did you get 7 and 9 from that?

7. precal

znimon is busy at the moment, have you tried to do factor by grouping

8. precal

|dw:1346462806136:dw|

9. precal

|dw:1346462837874:dw|

10. precal

|dw:1346462861944:dw|

11. precal

|dw:1346462941803:dw|

12. precal

|dw:1346462966127:dw|

13. precal

hope that helps

14. jim_thompson5910

Alternatively, you can set the expression equal to zero to get $\Large 4a^2-32a+63 = 0$ Now solve for 'a' using the quadratic formula $\Large a = \frac{-B\pm\sqrt{B^2-4AC}}{2A}$ $\Large a = \frac{-(-32)\pm\sqrt{(-32)^2-4(4)(63)}}{2(4)}$ $\Large a = \frac{32\pm\sqrt{1024-1008}}{8}$ $\Large a = \frac{32\pm\sqrt{16}}{8}$ $\Large a = \frac{32+\sqrt{16}}{8} \ \text{or} \ a = \frac{32-\sqrt{16}}{8}$ $\Large a = \frac{32+4}{8} \ \text{or} \ a = \frac{32-4}{8}$ $\Large a = \frac{36}{8} \ \text{or} \ a = \frac{28}{8}$ $\Large a = \frac{9}{2} \ \text{or} \ a = \frac{7}{2}$ Now rearrange to get the right sides equal to zero $\Large a = \frac{9}{2} \ \text{or} \ a = \frac{7}{2}$ $\Large 2a = 9 \ \text{or} \ 2a = 7$ $\Large 2a - 9 = 0 \ \text{or} \ 2a - 7 = 0$ Finally, use the zero product property to get $\Large (2a - 9)(2a - 7) = 0$ So this shows us that $\Large 4a^2-32a+63 = 0$ factors to $\Large (2a - 9)(2a - 7)$

15. lildancngurl