so the question is $(2x - 2)^2 + (3 - 2x)^2$. My working out: $$ 2x (3 - 2x)^2 + 2 (3 - 2x)^2 = 6x^2 - 4x + 6 - 4x = 2x^2 - 4x + 6. $$ I was a bit confused as their was another way to work out this question which is expanding and simplifying the first bracket separately to the second bracket and then later both join the two and then simplifying it...
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Your calculation looks strange. Use $$(a+b)^2=a^2+2ab+b^2.$$ Then, $$\begin{align}(2x-2)^2+(3-2x)^2&=(2x+(-2))^2+(3+(-2x))^2\\&=(2x)^2+2\cdot 2x\cdot (-2)+(-2)^2+3^2+2\cdot 3\cdot (-2x)+(-2x)^2\\&=4x^2-8x+4+9-12x+4x^2\\&=8x^2-20x+13.\end{align}$$
mathlove
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Instead of getting 12x I got 24x @mathlove – lase Oct 02 '14 at 11:04
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@lase: I have to say you are wrong. – mathlove Oct 02 '14 at 11:08
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I accidently added ^2. Ok I got -12x now – lase Oct 02 '14 at 11:09
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@lase: I see. Everyone does a mistake. No worry. – mathlove Oct 02 '14 at 11:10
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Based upon what I can see from your attempt so far, you seem to have become confused with the precedence order of applying the operations.
You need to treat each square separately (remember in order of precedence, the parentheses apply first), then apply the addition.
So expand $(2x-2)^2$ first, then $(3-2x)^2$, then add the two expansions together, and simplify the result.
If you get confused, remember to apply FOIL.
There are also two useful short cuts for squared sums and differences, that are worth committing to memory:
$(a+b)^2 = a^2 + 2ab + b^2$,
and:
$(a-b)^2 = a^2 - 2ab + b^2$
But be careful when you apply them, that you have to applied them to the coefficient too, so in your first expansion $(2x-2)^2$ the values are $a=2x$ and $b=2$, so $a^2=(2x)^2=4x^2$.
Hope that helps!
Chris Cooper
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