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Factor the trinomial completely.

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1 Answer

You have two slightly different polynomials showing.

y3 - 11y2 + 28y factors since you can pull out y to get y(y2 - 11y + 28).

Then noticing that -7 + (-4) = -11 and (-7)(-4) = 28, you get y(y - 4)(y - 7).

 

The other one you have is y3 - 11y2 + 28 which you can try the rational root test on.

The candidates are x = ±1, ±2, ±4, ±7, ±14, ±28. None of these turn out to be roots so it won't factor.

Another interesting problem is finding roots of any cubic polynomial. If you are curious on how to do that, you can check out my two-part video lesson.