Factor these polynomials. I really need to pass the 10th grade.?

Question

x to 2nd power + 6x + 9

4t to 2nd power + t - 3

2c to 2nd power - 5c - 3

x to 2nd power - 8x + 16

4m to 2nd power - 121

4g to 2nd power + 4g + 1

9t to 2nd power + 12t + 4

36s to 2nd power - 1

Answer 1

1)x^2+6x+9
=(x)^2+2*x*3+(3)^2
=(x+3)^2
2)4t^2+t-3
=4t^2+4t-3t-3
=4t(t+1)-3(t+1)
=(t+1)(4t-3)
3)2c^2-5c-3
= 2c^2+c-6c-3
=c(2c+1)-3(2c+1)
=(c-3)(2c+1)
4)x^2-8x+16
=(x)^2-2*x*4+(4)^2
=(x-4)^2
5)4m^2-121
=(2m)^2-(11)^2
=(2m+11)(2m-11)
6)4g^2+4g+1
=(2g)^2+2*2g*1+(1)^2
=(2g+1)^2
7)9t^2+12t+4
=(3t)^2+2*3t*2+(2)^2
=(3t+2)^2
8)36s^2-1
=(6s)^-(1)^2
=(6s+1)(6s-1)

Answer 2

#1 x to 2nd power + 6x + 9= (x+3)^2
#2 4t to 2nd power + t - 3= (4t+3)(t-1)
#3 2c to 2nd power - 5c - 3= (2c+1)(c-3)
#4 x to 2nd power - 8x + 16 = (x-4)^2
#5 4m to 2nd power - 121= (2m+11)(2m-11)
#6 4g to 2nd power + 4g + 1=(2g+1)^2
#7 9t to 2nd power + 12t + 4=(3t+2)^2
#8 36s to 2nd power - 1= (6s+1)(6s-1)

Answer 3

x² + 6x + 9 = (x+3)²
4t² + t - 3 = (4t-3)(t+1)
2c² - 5c - 3 = (2c+1)(c-3)
x² - 8x + 16 = (x-4)²
4m² - 121 = (2m+11)(2m-11)
4g² + 4g + 1 = (2g+1)²
9t² + 12t + 4 = (3t+2)²
36s² - 1 = (6s+1)(6s-1)

Answer 4

x^2 + 6x + 9
(x+3)(x+3)

look at the 9 how can you factor it?
9 * 1
3 * 3

notice 3+3 = 6 which is the middle term
so
(x+3)*(x+3)


4t^2 + t - 3

-3 = -1 * 3 or 1 * -3
the 4 on the front end can be 4 * 1 or 2 * 2
so you need a (+1) for the middle term

using 4* 1 and 1 * -3 you get 4-3 = 1 so

(4t - 3) (t + 1)

Better learn this.

Answer 5

1. (x+3) (x+3)

2. (4t -3) (t +1)

3. (2c+1) (c-3)

4. (x-4) (x-4)

5. (2m+11) (2m-11)

6. (2g+1) (2g+1)

7. (3t+2) (3t+2)

8. (6s+1) (6s-1)

Note: Factoring: What 2 values multiplied together give the
equation shown. You have to know how it's
solved, not just the answers.

Answer 6

PSHH im barely 13 years old and in 9th grade yet this is a piece of cake
1# (x+3)(x+3)
2# (t+1)(4t-3)
3# (c-3)(2c+1)
4# (x-4)(x+4)
5# (2m+11)(2m -11)
6# (2g+1)(2g+1)
7# (3t+2)(3t+2)
8# (6s+1)(6s -1

Answer 7

x to 2nd power + 6x + 9 = x^2+6x+9 = (x+3)(x+3) = (x+3)^2,

4t to 2nd power + t - 3 = 4t^2+t-3 = (2t+3)(t-1),

2c to 2nd power - 5c - 3 = 2c^2-5c-3 = (2c+1)(c-3)

x to 2nd power - 8x + 16 = x^2-8x+16 = (x-4)(x-4) = (x-4)^2,

4m to 2nd power - 121 = (2m+11)(2m-11)

4g to 2nd power + 4g + 1 = (2g+1)(2g+1) = (2g+1)^2

9t to 2nd power + 12t + 4 = (3t+2)(3t+2) = (3t+2)^2

36s to 2nd power - 1 = (6s+1)(6s-1)

Answer 8

x^2 + 6x + 9
(x + 3)(x + 3)

4t^2 + t - 3
(4t - 3)(t + 1)

2c^2 - 5c - 3
(2c + 1)(c - 3)

x^2 - 8x + 16
(x - 4)(x - 4)

4m^2 - 121
(2m + 11)(2m - 11)

4g^2 + 4g + 1
(2g + 1)(2g + 1)

9t^2 + 12t + 4
(3t + 2)(3t + 2)

36s^2 - 1
(6s + 1)(6s - 1)

Answer 9

(x +3) (x +3)
=(x + 3) to the power of 2

(4t -3) (t +1)

(2c +1) (c -3)

(x -4) (x -4)
=(x - 4) to the power of 2

(2m +11) (2m -11)

(2g +1) (2g +1)
= (2g +1) to the power of 2

(3t +2) (3t +2)
= (3t +2) to the power of 2

(6s +1) (6s +1)

Answer 10

If passing this try is the identifying ingredient for you transferring to the subsequent grade then I strongly propose you do your own artwork. in case you could not ascertain them out then ask your instructor, it truly is what they're there for. Math is a few thing that you're going to apply primary once you flow out into the international and make your mark.

Answer 11

1) (x + 3)(x + 3)
2) (4t - 3)(t + 1)
3) (2c + 1)(c - 3)
4) (x - 4)(x - 4)
5) (2m + 11)(2m - 11)
6) (2g + 1)(2g + 1)
7) (3t + 2)(3t + 2)
8) (6s + 1)(6s - 1)