Factor completely: 3x*4−6x*3+9x*2−18x

Question

The numbers after the stars are exponents. Thanks

Answer 1

3x⁴ - 6x³ + 9x² - 18x

 

All terms can be factored by 3 and by x

so start by factoring out 3x:

 

3x(x³ - 2x² + 3x - 6)

 

Now factor by grouping:

x³-2x² = x²(x-2)

3x-6 = 3(x-2)

 

so ... x³-2x²+3x-6 = (x-2)(x²+3)

 

Final answer:

3x(x-2)(x² + 3)

Answer 2

3x(x-2)(x²+3) by the use of group factoring.  Please learn how to show us an exponent, e.g. x^2 or x².

Answer 3

3x*4−6x*3+9x*2−18x

 

First, factor out the GCF: 3x(x³ - 2x² + 3x - 6)

 

Group the expression in the parenthesis in twos.

 

Thus, we have 3x[(x³ - 2x²) + (3x - 6)]

 

Factor these groups. Then we have, 3x[x²(x - 2) + 3(x - 2)]

 

Factor the expression again since (x-2) is common to both groups:

 

3x[(x² + 3)(x - 2)]

 

You can check your answer by multiplying the factors.

Answer 4

Assuming this is 3x⁴ - 6x³ + 9x² -18x

 

We need to find a greatest common factor for all the terms (and also keep the leading term positive).

That factor is 3x (since it can divide all the terms evenly).

 

So, we'll pull out that factor and divide the other terms by it:

 

3x(x³ - 2x² + 3x - 6)

 

Now factor the expression within the parentheses (save the 3x for the end):

1. Grouping: x³ - 2x² + 3x - 6 = (x³ - 2x²) + (3x - 6)

2. Factor each group: x²(x - 2) + 3(x - 2)  Notice that both terms now have a common factor: x - 2

3. Factor out the common factor: (x - 2)(x² + 3)

4. Bring all 3 factors together (including the 3x from the beginning): 3x(x - 2)(x² + 3)