(1) Factor `-1/4(2x)^2+25(2y^3)^2`

First rewrite as `25(2y^3)^2-1/4(2x)^2`

Then `5^2(2y^3)^2-(1/2)^2(2x)^2`

Now `(10y^3)^2-(1x)^2` `a^2*b^2=(a*b)^2`

We have the difference of two squares: `u^2-v^2=(u+v)(u-v)`

So `(10y^3+x)(10y^3-x)`

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**Thus the factors are `10y^3+x,10y^3-x` **

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We can check by multiplying:

`(10y^3+x)(10y^3-x)`

`=(10y^3)(10y^3)-10y^3x+10y^3x-x^2`

`=(10y^3)^2-(x)^2`

`=(5*2y^3)^2-(1/2*2x)^2`

`=5^2(2y^3)^2-(1/2)^2(2x)^2`

`=25(2y^3)^2-1/4(2x)^2`

`=-1/4(2x)^2+25(2y^3)^2` as required.

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(2) The product of...

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(1) Factor `-1/4(2x)^2+25(2y^3)^2`

First rewrite as `25(2y^3)^2-1/4(2x)^2`

Then `5^2(2y^3)^2-(1/2)^2(2x)^2`

Now `(10y^3)^2-(1x)^2` `a^2*b^2=(a*b)^2`

We have the difference of two squares: `u^2-v^2=(u+v)(u-v)`

So `(10y^3+x)(10y^3-x)`

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**Thus the factors are `10y^3+x,10y^3-x` **

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We can check by multiplying:

`(10y^3+x)(10y^3-x)`

`=(10y^3)(10y^3)-10y^3x+10y^3x-x^2`

`=(10y^3)^2-(x)^2`

`=(5*2y^3)^2-(1/2*2x)^2`

`=5^2(2y^3)^2-(1/2)^2(2x)^2`

`=25(2y^3)^2-1/4(2x)^2`

`=-1/4(2x)^2+25(2y^3)^2` as required.

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(2) The product of two consecutive even integers is 16 times more than 8 times the smaller. Find the integers.

Let 2x,2x+2 be the integers. (They are even since both are a multiple of 2, and they are consecutive since they are 2 units apart.)

Then (2x)(2x+2)=16(8(2x))

`4x^2+4x=256x`

`4x^2-252x=0`

`(4x)(x-63)=0`

`=>x=0` or `x=63`

Both of these are solutions. `0*2=16(8(0))=0` . Also if x=63, then 2x=126 and `126(128)=16(8(126))=16128`

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**Thus the numbers we seek are 126 and 128.**

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