Could you please help me with these factoring quadratic trinomial problems?

Question

Please please help me. I would really appreciate it.
Thanks in advance.

1.) y squared + 21- 72

2.) c squared + 16c-80

3.) d squared + 13d +40

4.) y squared + 21y -72

5.) 3a squared +10a +7

6.) 5 n squared + 4n -12

7.) 6 y squared + 109y +18

8.)14 n squared + 17n -18

9.) 2x squared - 5xy - 3y squared

10.) 8 a squared + 22ab +15 b squared

11.) 6 x squared + 5x squared y squared + y (to the fourth power)

Answer 1

i dont quite remember how to do it exactly. but i know its reverse "FOIL". i dont know if you learned "FOIL" but your answer should look like (x-6)(y+4)....or something along those lines. is teh first one a typo? maybe y^2 +21y-72..but that makes it the same as #4. the only thing i cant remember is where the number would have to go in the parenthesis answer to be correct. i think its usually the variable (x, y, or whatever) and then the number so that you get an answer like i showed you before. gosh, i really miss math. check the beginning section of that chapter in your book. it should explain it. good luck

Answer 2

wow. you really need to learn this and be able to do it on your own.. i will start you out.

with no coefficients in front of the first term, the backbone will be: ( y + ? ) ( y + ? ) --- or whatever variable..
the key is to look for multiples of the 3rd term that add up to the 2nd term...
Let's look at #1:
y^2 + 21 - 72
Backbone: (y + ?)(y + ?)
Multiples of 72: 9 and 8; 36 and 2; 3 and 24.
Multiples that add up to 21: -3 and +24:
Giving us: (y - 3)(y + 24)...

Those with a coefficient in front of the first term are trickier and just take practice.. You do the same thing, but it is just more involved and I just do more guess and checking...
Let's look at #5:
3a^2 + 10a + 7
Now you have to consider multiples of the first term to set that up first. Here, 3 is prime so we know there is only one possibility:
(3a + ?)(a + ?).
Now we have to look at multiples of 7 - again prime: 7 and 1.
Then we have to arrange the 1 and 7 to work out in the equation to give us 10a:
(3a + 7)(a + 1)
Check:
First: 3a*a = 3a^2
Outside: 3a * 1 = 3a
Inside: 7*a = 7a
Last: 7*1= 7
Add like terms:
3a^2 + (3a + 7a) + 7 =
3a^2 + 10a + 7.
boom.

Answer 3

1) (y+24)(y-3)
2) (c+20)(c-4)
3) (d+8)(d+5)
4) (y+24)(y-3) [same as #1???]
5) (3a+7)(a+1)
6) (5n-6)(n+2)
7) (6y+1)(y+18)
8) (14n-9)(n+2)
9) (2x+y)(x-3y)
10) (2a+3b)(4a+5b)
11) I'm assuming the 6x squared should be 6x to the fourth?? If yes, then the answer is, (2x squared + y squared)(3x squared + y squared).

If the 5xy number is actually 5(x)(y squared) instead of 5(x squared)(y squared), then the answer is (2x + y squared)(3x + y squared)

Answer 4

1.
y^2 + 21 - 72
y^2 - 72
(y - sqrt72)(y + sqrt72)

2.
c^2 + 16c - 80
(c + 20)(c - 4)

3.
d^2 + 13d +40
(d + 5)(d + 8)

4.
y^2 + 21y - 72
???

5.
(3a^2 + 10a + 7
(3a + 7)(a + 1)

6.
5n^2 + 4n -12
(5n - 6)(n + 2)

7.
6y^2 + 109y + 18
(6y + 1)(y + 18)

8.
14n^2 + 17n -18
????

9.
2x^2 -5xy -3y^2
(2x +y)(x - 3y)

10.
8a^2 + 22ab +15b^2
(4a +5b)(2a + 3b)

11.
6x^2 + 5x^2y^2 + y^4
???? I think you mean 6x^4
6x^4 + 5x^2y^2 + y^4
(3x^2 + y^2)(2x^2 + y^2)