How do you factor #3j^3-51j^2+210j#?

Answer 1

f(j) = 3j(j - 7)(j - 10)

#f(j) = 3j(y) = 3j(j^2 - 17j + 70)# Factor y by the new AC Method (Socratic Search). Find two numbers knowing sum (-17= b) and product (c = 70). These numbers have same sing because ac > 0 Factor pairs of (70) --> (7, 10)(-7, -10). This sum is (-17 = b). Therefor, y = (j - 7)(j - 10) f(j) = 3j(j - 7)(j - 10)

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Answer 2

Jameson Allen

To factor the expression (3j^3 - 51j^2 + 210j), you can first factor out the greatest common factor, which is (3j).

[3j(j^2 - 17j + 70)]

Now, factor the quadratic expression inside the parentheses. Find two numbers that multiply to (70) and add up to (-17). These numbers are (-7) and (-10).

[3j(j^2 - 10j - 7j + 70)]

Now, factor by grouping:

[3j(j(j - 10) - 7(j - 10))]

Factor out the common factor (j - 10):

[3j(j - 10)(j - 7)]

So, (3j^3 - 51j^2 + 210j) factors to (3j(j - 10)(j - 7)).

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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.