How do you factor completely #x^3 - 8x^2 + 5x + 50 = 0#?

Answer 1

#(x+2)(x-5)^2#.

We know that if a number solves such an equation, it must divide its last coefficient, i.e. #50#. So, we can try with some of its divisors: if #f(x)=x^3-8x^2+5x+50#, then:

So, #-2# is a solution, which means that #f(x)# can be divided by #(x+2)#. Do the long division, and you have that

#x^3-8x^2+5x+50 = (x+2)(x^2-10x+25)#

Now we must factor the second parenthesis, but you can see that it is the square of #(x-5)#, and so #f(x)# is completely factored.

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Answer 2

Julia Adams

To factor completely the equation x^3 - 8x^2 + 5x + 50 = 0, you can use synthetic division or factor by grouping. After factoring, the equation can be expressed as:

(x - 5)(x^2 - 3x - 10) = 0

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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.