I found a question in the textbook that I'm struggling to answer.

Simplify this expression by expanding brackets, factorising or both:

[(2x+y)(y-2x)]²-(4x²+y²+2z²)²

At the back of the book with the answer it gives the advice that it is the difference of two squares, so I attempted to do it like that but I got a different answer from the textbook.

The answer in the textbook is -4(4x²+z²)(y²+z²)

Can someone please help explain where I went wrong I will post my working out

Simplify this expression by expanding brackets, factorising or both:

[(2x+y)(y-2x)]²-(4x²+y²+2z²)²

At the back of the book with the answer it gives the advice that it is the difference of two squares, so I attempted to do it like that but I got a different answer from the textbook.

The answer in the textbook is -4(4x²+z²)(y²+z²)

Can someone please help explain where I went wrong I will post my working out

Original post by JimMiller069

I found a question in the textbook that I'm struggling to answer.

Simplify this expression by expanding brackets, factorising or both:

[(2x+y)(y-2x)]²-(4x²+y²+2z²)²

At the back of the book with the answer it gives the advice that it is the difference of two squares, so I attempted to do it like that but I got a different answer from the textbook.

The answer in the textbook is -4(4x²+z²)(y²+z²)

Can someone please help explain where I went wrong I will post my working out

Simplify this expression by expanding brackets, factorising or both:

[(2x+y)(y-2x)]²-(4x²+y²+2z²)²

At the back of the book with the answer it gives the advice that it is the difference of two squares, so I attempted to do it like that but I got a different answer from the textbook.

The answer in the textbook is -4(4x²+z²)(y²+z²)

Can someone please help explain where I went wrong I will post my working out

So this is how I worked out my answer:

[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]

[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)(-y²-z²)

Original post by JimMiller069

So this is how I worked out my answer:

[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]

[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)(-y²-z²)

[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]

[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)(-y²-z²)

I forgot to type two square brackets in the middle but I did include them

Original post by JimMiller069

So this is how I worked out my answer:

[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]

[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)(-y²-z²)

[((2x+y)(y-2x))-(4x²+y²+2z²)][((2x+y)(y-2x))+(4x²+y²+2z²)]

[(y²-4x²)-(4x²+y²+2z²)(y²-4x²)+(4x²+y²+2z²)]

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)(-y²-z²)

Youre not miles off, but Id have done dots on

((2x+y)(y-2x))

first, then dots again, which youve sort of done. But at the end you have both a 2 and a -2 factor which gives -4 in front of the remaining expression.

(edited 1 month ago)

Original post by mqb2766

Youre not miles off, but Id have done dots on

((2x+y)(y-2x))

first, then dots again, which youve sort of done. But at the end you have both a 2 and a -2 factor which gives -4 in front of the remaining expression.

((2x+y)(y-2x))

first, then dots again, which youve sort of done. But at the end you have both a 2 and a -2 factor which gives -4 in front of the remaining expression.

Oh so it's

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)2(y²+z²)

Then -2x2 is -4 so -4(4x²+z²)(y²+z²)?

Original post by JimMiller069

Oh so it's

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)2(y²+z²)

Then -2x2 is -4 so -4(4x²+z²)(y²+z²)?

(-8x²-2z²)(2y²+2z²)

-2(4x²+z²)2(y²+z²)

Then -2x2 is -4 so -4(4x²+z²)(y²+z²)?

Yes. Algebraically if you had

2ab

you could interpret that as either

(2a)b or a(2b)

so if you have (2a)(2b) then its 4ab

Original post by mqb2766

Yes. Algebraically if you had

2ab

you could interpret that as either

(2a)b or a(2b)

so if you have (2a)(2b) then its 4ab

2ab

you could interpret that as either

(2a)b or a(2b)

so if you have (2a)(2b) then its 4ab

Thank you for your help!!

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