Algebra Help?

Question

Perform the indicated operations and simplify:

(2y)2 (2y3)


x2 + 13x +40 = 0


3x3 – 588x = 0


x2 + 2x = 63


49p2 – 112p + 64 = 0

Factor this trinomial completely: 21y2 + y – 2



THANKS FOR ALL THE HELP!!!!!

Answer 1

1)
(2y)2 (2y3)
= 4y(2y^3)
= 8y^4

2)
x2 + 13x +40 = 0
x^2 + 5x + 8x + 40 = 0
x(x + 5) + 8(x + 5) = 0
(x + 8)(x + 5) = 0
x = -8
or
x = -5

3)
3x^3 – 588x = 0
3x(x^2 - 196) = 0
3x(x-14)(x+14) = 0
x = 0
or
x = 14
or
x = -14

4)
x^2 + 2x = 63
x^2 + 2x - 63 = 0
x^2 - 7x + 9x - 63 = 0
x(x-7) + 9(x-7) = 0
(x + 9)(x - 7) = 0
x = -9
or
x = 7

5)
49p2 – 112p + 64 = 0
(7p)^2 - 2(7p)(8) + (8)^2 = 0
(7p - 8)^2 = 0
p = 8/7

6)
21y2 + y – 2
=21y^2 - 7y + 6y - 2
=7y(3y - 1) + 2(3y - 1)
= (7y + 2)(3y - 1)

Answer 2

(2y)2 (2y3) =4y^5

x2 + 13x +40 = 0
(x+8)(x+5) =0 --> x = -5 or -8

3x3 – 588x = 0
3x(x^2-196) = 0 --> x=0 or -14 or +14

x2 + 2x = 63
(x+9)(x-7) = 0 ==> x = -9 or 7

49p2 – 112p + 64 = 0
(7p-8)^2 = 0
7p-8 = 0 --> p = 8/7

Factor this trinomial completely: 21y2 + y – 2
(3y+1)(7y-2)

Answer 3

1) Not sure if those are exponents or not...
If it is (2y)^2 * 2y^3, then the answer is 8y^5
2) (x+5)(x+8)
x = -5, -8

3) 3x ( x^2 - 196) <-- factoring out the 3x
3x ( x - 14) ( x+14) <-- difference of squares
x = 0, 14, -14

4. x^2 + 2x - 63 = 0
(x-7)(x+9)
x=7, -9

5. (7p-8)(7p-8) or (7p-8)^2

p = 8/7

6. (7y-2)(3y+1)

Answer 4

x2+13x+40 = (x+5)(X+8) X= -5,-8

3x3-588x=0 equals 3x(x2-196) so x= 0,14,-14

x2+2x=63 equals (x-7)(x+9) x= 7, -9

49p2-112p+64 is (7p-8) squared p=8/7

trinomial = (3y+1) (7y-2) y= -1/3, 2/7

Answer 5

1)
8y^5
2)
(x + 5)(x + 8) = 0
x =-5 or -8
3)
3x(x^2 - 196) = 0
3x(x + 14)(x - 14) = 0
x = 0, 14 , -14
4)
x^2 + 2x - 63 = 0
(x + 9)(x - 7) =0
x = -9 or 7
5)
49p^2 - 112p + 64 =0
(7p - 8)^2 = 0
p = 8/7
6)
21y^2 + 7y - 6y -2 =0
7y(3y + 1) -2(3y + 1) =0
(3y + 1)(7y - 2) =0
y = -1/3 or 2/7

Answer 6

1) (2y)2(2y3) = 96*4y
I'll use commutative and distributive property
2(2y)(2y*3)
2(2)*2y*2(2)*2y*2(3)
4*2y*4*2y*6
Multiply variables with variables and normal numbers with normal numbers.
4*4*6*2y*2y
96*4y

2) x2 + 13x +40 = 0
2x + 13x + 40 = 0
15x + 40 = 0
15x = 0 - 40
15x = -40
x= -40/15 = -2.6

3) Couldn't do it, sorry =S

4) x2 + 2x = 63
2x + 2x = 63
4x = 63
x = 63/4 = 15.75

5) Couldn't do it =/

Answer 7

2y5

(x+5)(x+8)=0 (x=-5, -8)

(3x + 42)(x - 14) = 0, x= 14

x2 + 2x - 63 = 0, (x+9)(x-7) = 0, x = -9, 7

(7p + 8)(7p + 8) = 0, x= -8/7

(3y+1)(7y-2)

Answer 8

8y^5
(x+8)(x+5)=0 x=-8, x=-5
3x(x^2-196)=0 x=0, x= +/- 14
(x+9)(x-7)=0 x=-9, x=7
(7p -8)(7p-8) p=8/7
(7y-2)(3y+1)