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# 2*(x+4)-((1-(3*x))^2)

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2*(x+4)-((1-(3*x))^2)
Guest Jan 11, 2015
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#1
+11843
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First of all what was the need to use do many brackets! your question is looking more like a shop of brackets or in other words like Einsteins hair! Anyways I'll solve it

2*(x+4)-((1-(3*x))^2)

now multiply 2 by x and 4 , multiply 3 by x and just cut all those useless brackets!

= 2x + 8 - (1- 3x)^2

now here we will use an identity for the expression in brackets

Identity is (a-b)^2= a^2 - 2ab + b^2

now we will use this identity to solve the bracketed expression!

= 2x + 8 - 1 - 2x1x3x + (3x)^2

= 2x + 8 - 6x + 9x^2

now we wil arrange them with their alikes

= 2x - 6x + 8 + 9x^2

= -4x + 8 + 9x^2  ANSWER

rosala  Jan 11, 2015
#2
+91900
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2*(x+4)-((1-(3*x))^2)

\$\$\\(1-(3*x))^2\\
=(1-3x)^2\\
=1-6x+9x^2\\\$\$

2(x+4)-(1-6x+9x2)

=2x+8-1+6x-9x2

= -9x2+8x+7

Melody  Jan 11, 2015
#3
+91900
0

Can you work out who got what wrong?

Melody  Jan 11, 2015
#4
+808
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The correct answer is: -9x2+8x+7   (checked by wolfram alpha)

I think the wrong answer originated from here:

"... 2x + 8 - (1- 3x)^2

now here we will use an identity for the expression in brackets

Identity is (a-b)^2= a^2 - 2ab + b^2

now we will use this identity to solve the bracketed expression!

= 2x + 8 - 1 - 2x1x3x + (3x)^2 ..."

I'll solve it like this:

2x + 8 - (1 - 3x)2 =

2x + 8 - (1 - 6x + 9x2) =

2x + 8 - 1 + 6x - 9x2

So it's mistakes made based on the minus before the paranthesis.

Tetration  Jan 11, 2015
#5
+91900
0

Thanks Tetration. :)

Melody  Jan 11, 2015

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