Just tell me the answer

x(x+1)(x+2)

simply multiply the first bracket with x and then with the total multiply it with the second bracket

x^{2}+x(x+2)

cross multiply:

x^{3}+2x^{2}+x^{2}+2x

x^{3}+3x^{2}+2x

Just tell me the answer

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Puja K. | good in solving algebra problemsgood in solving algebra problems

x(x+1)(x+2)

simply multiply the first bracket with x and then with the total multiply it with the second bracket

x^{2}+x(x+2)

cross multiply:

x^{3}+2x^{2}+x^{2}+2x

x^{3}+3x^{2}+2x

Go for a longer but the simplest method: x(x+1)(x+2)

x(x+1)=(x^{2}+x)

(x^{2}+x)(x+2)=x^{3}+3x^{2}+2x

Not sure if this is a question on finding the roots, if it is:

You look to see what x values can make the whole expression zero. Well all those three quantities (the three sets of parenthesis), are all multiplying together.

x * (x+1) * (x+2). Well the only ways to make something times something equal zero is what... think about that for a second before you read...

if one of the something's themselves is zero (8 * 0 = 0 for example).

With that in mind, what makes the first quantity (x) equal zero? What value makes the second quantity (x+1), equal zero? and the third quantity? Those will be your three options to making the expression as a whole equal zero.

Jessica M. | Mathematics AssistanceMathematics Assistance

x (x+1) (x+2)=

First distribute the x:

(x^{2}+x) (x+2)=

Then use the foil method:

(x^{2}*x + x^{2}*2 + x*x + x*2)=

Finally collect the terms and you have:

**(x ^{3}+3x^{2}+2x)**

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