Factorise (x^3)-8?

Answer 1

You are able to do long division with polynomial expresions just like normal numbers.

If you take each term and treat it like the hundreds digit, the tens digit, the ones digit, etc..., you can factor certain expressions a little more easily.

The expression (x^3)-8 is actually:
(x^3) + 0*(x^2) + 0*(x) + (-8)

Just like 1,432 is actually:
1000*(1) + 100*(4) + 10*(3) + 1*(2)

For factoring a polynomial expression, try to find something the FIRST term and the LAST term have in common.

In this case, the are both CUBES of whole numbers, the first term is x cubed, and the last term is -2 cubed.

If you perform long division on (x^3) - 8 using the divisor x-2 (or x +(-2)), you will end up with the quotient (x^2) + 2x + 4, which can not be simplified.

Therefore,

(x^3)-8 = (x-2) * ((x^2) + 2x + 4)

If you would like to SEE an example of long division with polynomial expressions (which really DOES help with understanding it) you can e-mail me through my Yahoo answers profile.

I hope this helped.

Answer 2

Use the "difference of cubes" rule. The first and last terms are easy:
(x - 2)(x^2.....+4)

But if that were all, you would have middle terms of -2x^2 and 4x. Those will cancel out if you add:
(x - 2)(x^2+2x+4)

Easiest to explain if you are familiar with the 'box foil' method.

Answer 3

useing this formula
a^3 - b^3 = (a-b) (a^2-ab+b^2 )

so that

x^3 - 8 = ( x )^3 - ( 2 )^3
= (x - 2 ) ( x^2 - 2x +2^2 )
= ( x-2 ) ( x^2 - 2x + 4 )

Solved

Raghavkirti

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Answer 4

(x-2) (x^2 + 2x + 4)