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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.

I have added stuff  that might help you a bit.

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

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

Your answer appears to bre perfect Sasini.  Thank you

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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