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# What is the solution to the equation 1−5(x3+x2)=-5x2−4(x3+2)?

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What is the solution to the equation 15(x3+x2)=-5x24(x3+2)?

Dec 23, 2014

#5
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i know i am sorry ok just trying

Dec 23, 2014

#1
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When you write these Shaniab you should use the 'hat' to indicate exponents. :)

1−5(x^3+x^2)=-5x^2−4(x^3+2)

I would like you to try and do it yourself Shaniab

Just remember that the minus sign belongs to the number that follows it.

1+[−5(x^3+x^2)]=-5x^2+[−4(x^3+2)]

Now can you give it a go?

Dec 23, 2014
#2
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15(x3+x2)=-5x24(x3+2)15x35x2=-5x24x3815x2=-5x2+x3895x2=-5x2+x39=x327=x

Dec 23, 2014
#3
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1 -5(x^3+x^2) = -5x^2 -4(x^3 +2)

1 -5x^3 -5x^2  = -5x^2 -4x^3 -8

1 -5x^3            =  -4x^3 - 8

1 +8                 =   -4x^3  + 5x^3

9                      =      x^3

Dec 23, 2014
#4
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So

$$x=\sqrt[3]{9}$$

Shaniab, It is important that when you show woking for your answers that you only ever put one equal sign per line.

It is too hard to think it through otherwise :)

Dec 23, 2014
#5
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