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Ditto..

actor the following: 25 a^4+40 a^2+16 Using 25 a^4+40 a^2+16 = (5 a^2)^2+2 (5 a^2) (4)+(4)^2, the coefficients match the 3^rd row of Pascal's triangle, so 25 a^4+40 a^2+16 = (5 a^2+4)^2: Answer: | | (5a^2+4)^2

25a^4+40a^2+16

25a⁴ + 40a² + 16 = (5a² + 4)² 

Perform the indicated operations on the following polynomials. 

2.Divide: 6x3 + 27x - 19x2 - 15 by 3x -5 

..............2x² – 3x + 4 
...........__________________ 
3x – 5 ) 6x³ – 19x² + 27x – 15 
..............6x³ – 10x² 
............--------------------------... 
....................– 9x² + 27x – 15 
....................– 9x² + 15x 
...................-------------------... 
................................12x – 15 

This was also on my homework.. :D


(5a³ – 4a + 7) – (6a³ – 2a² + 4) = 5a³ – 4a + 7 – 6a³ + 2a² – 4) 
......................................... = 5a³ – 6a³ + 2a² – 4a + 7 – 4 
......................................... = – a³ + 2a² – 4a + 3 


4.Factor the following polynomial completely. 

y² + 12y + 36 = (y + 6)²

Me too 

I agree with CPhil:)

Let's go with the bad, first.......it's always better to end up feeling good.....!!!!

Yupp :p

  20x2 + 22xy + 6y2 = 2(10x2 + 11xy + 3y2) = 2(5x+3y)(2x+y)

Factor the following: 20 x^2+22 x y+6 y^2 Factor 2 out of 20 x^2+22 x y+6 y^2: 2 (10 x^2+11 x y+3 y^2) The coefficient of x^2 is 10 and the coefficient of y^2 is 3. The product of 10 and 3 is 30. The factors of 30 which sum to 11 are 5 and 6. So 10 x^2+11 x y+3 y^2 = 10 x^2+6 x y+5 x y+3 y^2 = 5 x (2 x+y)+3 y (2 x+y): 2 5 x (2 x+y)+3 y (2 x+y) Factor 2 x+y from 5 x (2 x+y)+3 y (2 x+y): Answer: | | 2 (2 x+y) (5 x+3 y)

Answer :  A variable

well this is under off topic catagory 

real  sliopknot fans know that slipnkot only has FOUR song...^{. looser}

real  sliopknot fans know that slipnkot only has FOUR song...^{. looser}

Solve the following system: {2 x+3 y = 0 | (equation 1) x+2 y = -1 | (equation 2) Subtract 1/2 × (equation 1) from equation 2: {2 x+3 y = 0 | (equation 1) 0 x+y/2 = -1 | (equation 2) Multiply equation 2 by 2: {2 x+3 y = 0 | (equation 1) 0 x+y = -2 | (equation 2) Subtract 3 × (equation 2) from equation 1: {2 x+0 y = 6 | (equation 1) 0 x+y = -2 | (equation 2) Divide equation 1 by 2: {x+0 y = 3 | (equation 1) 0 x+y = -2 | (equation 2) Collect results: Answer: | | {x = 3              y = -2

2x + 3y = 0

x + 2y = -1

Notice that if we multiply the second equation by -2 all the way through, we get

-2x  - 4y = 2      

So we have

2x + 3y = 0

-2x  - 4y = 2        add these and we get

       -1y  = 2          divide both sides by -1

y  = -2      and using    x + 2y = -1     we have that  

x + 2(-2)  = -1 

x  -  4  = -1         add 4 to both sides

x = 3

So   (x , y)  = (3, -2)

Injusta, I believe he needs x.

Factor the following: 96 x^2-6 y^2 Factor 6 out of 96 x^2-6 y^2: 6 (16 x^2-y^2) 16 x^2-y^2 = (4 x)^2-y^2: 6 ((4 x)^2-y^2) Factor the difference of two squares. (4 x)^2-y^2 = (4 x-y) (4 x+y): Answer: | | 6 (4 x-y) (4 x+y)

im the op.

thanks a lot man much appreciated.

To solve x or y?

96x^2-6y^2

Factoring out 6 from (96x2−6y2), we get:

6(16x2−y2)

Factoring the equation above:

6(4x−y)(4x+y)→ final answer

Checking using FOIL method:

(4x−y)(4x+y)=16x2+4xy−4xy−y2=16x2−y2

6(16x2−y2)=(96x2−6y2)

Did I do it right???

1. Get off internet.

2. Ask real people.

3. Continue conversation from there.

"All the way....

  With JQA"

square root of 273.64

16.5420675854

Good, I wasn't just makin a stupid mistake...

That was one big step.

My step is go on ebay