How do you factor completely #2x^3+10x^2+14x+70#?

Answer 1

#2x^3+10x^2+14x+70 = 2(x^2+7)(x+5)#

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping, but first we can separate out the common scalar factor #2#...

#2x^3+10x^2+14x+70 = 2(x^3+5x^2+7x+35)#

#color(white)(2x^3+10x^2+14x+70) = 2((x^3+5x^2)+(7x+35))#

#color(white)(2x^3+10x^2+14x+70) = 2(x^2(x+5)+7(x+5))#

#color(white)(2x^3+10x^2+14x+70) = 2(x^2+7)(x+5)#

That's as far as we can go with Real coefficients.

If you allow Complex coefficients then:

#x^2+7 = (x-sqrt(7)i)(x+sqrt(7)i)#

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Answer 2

Kennedy Adams

To factor completely (2x^3 + 10x^2 + 14x + 70), you can first look for the greatest common factor, which is (2). Then, you can factor out (2) from each term:

[2(x^3 + 5x^2 + 7x + 35)]

Next, you can try to factor the expression (x^3 + 5x^2 + 7x + 35). However, it does not factor further using integer coefficients. Therefore, the factored form of (2x^3 + 10x^2 + 14x + 70) is (2(x^3 + 5x^2 + 7x + 35)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.