help with factoring?

Question

can you help me factor some math problems?

1.7a^2+22a+3

2.3a^2+14a+15

3.2y^2-7y-2

4.3y^2+5y-2

5.3m^2-7m-6

6.4b^2+5b-6

7.2x^2+5x-12

8.7n^2-22n+3

9.3x^2+4x-15

10.6y^2-11y+14

11.4y^2-17y-15

12.10n^2-19n+7

13.9k^2-12k+4

14.5b^2-13b-10

15.3t^2-32t+20

If you are going to say do your own homework, please do all of us a favor and save it. If i knew how to do this, I would not be asking for help!

it is still Thursday....it will be Friday in 10 minutes though....

Answer 1

1.7a^2+22a+3
(7a+1)(a+3)

2.3a^2+14a+15
(3a+5)(a+3)

3.2y^2-7y-2
y=[7+/-rt(49+16)]4

4.3y^2+5y-2
(3y+1)(y+2)

5.3m^2-7m-6
(3m+2)(m-3)

6.4b^2+5b-6
(4b-3)(b+2)

7.2x^2+5x-12
(2x-3)(x+4)

8.7n^2-22n+3
(7n-1)(n-3)

9.3x^2+4x-15
(3x-5)(x+3)

10.6y^2-11y+14
y=[11+/-rt(121-336)]/12

11.4y^2-17y-15
(4y+3)(y-5)

12.10n^2-19n+7
(5n-7)(2n-1)

13.9k^2-12k+4
(3k-2)^2

14.5b^2-13b-10
b=[13+/-rt(169+200)]/10

15.3t^2-32t+20
t=[32+/-(1024-240)]/6

Answer 2

1.7a^2+22a+3
(7a+1)(a+3)

2.3a^2+14a+15
(3a+5)(a+3)

3.2y^2-7y-2
Has no perfectsquare. Use quadratic equation

4.3y^2+5y-2
(3y-1)(y+2)

5.3m^2-7m-6
(3m+2)(m-3)

6.4b^2+5b-6
(4b-3)(b+2)

7.2x^2+5x-12
(2x-3)(x+4)

8.7n^2-22n+3
(7n-1)(n-3)

9.3x^2+4x-15
(3x-5)(x+3)

10.6y^2-11y+14
If the equation were 6y^2-11x-14, the factors would be (3x-14)(2x+1). OOps, I used x's instead of y's. anyway, since it is +14, I am not sure if it has roots unless you use the quadratic equation.

11.4y^2-17y-15
(4y+3)(y-5)

12.10n^2-19n+7
(2n-1)(5n-7)

13.9k^2-12k+4
(3k-2)(3k-2)

14.5b^2-13b-10
???????

15.3t^2-32t+20
(3t-2)(t-10)

That is a lot. You should learn how to do these things.

Answer 3

1 (7a+1 )(a+3 )
2 (3a+5 )(a+3 )
3 Will not factor.
4 Will not factor.
5 (3m+2 )(m-3 )
6. (4b-3)(b+2)

7. (x+4)(2x-3)

8. (7n-1)(n-3)

9. (3x-5)(x+3)

10. Will not factor.

11. (4y+3)(y-5)

12. (5n-7)(2n-1)

13. (3k-2)(3k-2)

14. will not factor.

15. (3t-2)(t-10)

I would NEVER SAY THAT!! I would like the opportunity to explain these to you sometime when we both have the time. I am Eds and I understand the "PROBLEM". Have a great Friday!!
Eds

PS The person "above me" said a= -3. It is NOT... These are not equations. JUST simple factorizations. Have a great holiday break.

Answer 4

i'll help you out a little, but i would really talk to my teacher about some extra help because you might want her/him to explain it again for you.

lets use #1 as an example.
you start by doing the "magic method" and what you do is multiply the first coeficient into the last number
so in #1 you have:
7a^2+22a+3
so your first coeficient is 7 and your last number is 3
so your new equation is a^2 + 22a + 21
now what you do is find the two answers which would be (a+1) and (a+21) (--- and if you dont know how to do this then i would REALLY talk to your teacher because you wont understand anything else)

BUT you're not finished. since you did the magic method, you need to take that first coeficient, and put it BACK in both answers.
so you have (7a+1) and (7a+21)
then you set those equal to zero, and you get that a = - 1/7 and
a= - 21/7
which is reduced to a = - 3
done.