# Video: Solving Polynomial Inequalities

Solve the inequality πβ΄ β πΒ³ β 20πΒ² < 0.

05:48

### Video Transcript

Solve the inequality π to the fourth power minus π cubed minus 20π squared is less than zero.

Now the first thing we notice is that this is a quartic inequality, as the highest power of π which appears is four. However, as we donβt have a term in π or the constant term, we know that we can actually factor the left-hand side of this inequality by π squared. Doing so gives π squared multiplied by π squared minus π minus 20 is less than zero. And so we have a quadratic term, π squared, multiplied by a quadratic expression, π squared minus π minus 20. And this must all be less than zero.

Now letβs consider our two factors separately. And weβll begin by thinking about π squared. Now π squared, by its very definition, must be greater than or equal to zero. Whenever we take a value, whether itβs positive or negative, and square it, we always get an answer which is either positive or equal to zero if the value we were squaring was itself zero. So what weβre doing is taking something which is greater than or equal to zero, multiplying it by a quadratic expression, another value. And we want to get an answer which is less than zero.

The only way this can happen is if two things are true. The quadratic π squared minus π minus 20 must itself be less than zero. But it must also be true that π is not equal to zero. Because if π were equal to zero, weβd have zero squared multiplied by this quadratic expression, which would give zero. Thatβs a value equal to zero, not less than zero.

So if we ensure that π is not equal to zero, π squared will then be strictly greater than zero. So we have a positive value multiplied by the parentheses. The parentheses must therefore be negative so that we have positive multiplied by negative, which gives negative overall. So weβre therefore looking to solve what is now a quadratic inequality. π squared minus π minus 20 is less than zero. But we must also remember that π cannot be equal to zero. So if zero is part of the solution set for the quadratic inequality, we must remember to take this individual value out.

So weβre now looking to solve this quadratic inequality. And the first step is to see whether it can be factored. As the coefficient of π squared is one, the first term in each of our parentheses will simply be π. As π multiplied by π gives π squared. To complete our parentheses, weβre then looking for two values whose sum is the coefficient of π. Thatβs negative one. And whose product is the constant term. Thatβs negative 20.

With a little bit of thought, we see that the two numbers which have both of these properties are four and negative five. So we can include these as the second term in each of our parentheses. And we have that π plus four multiplied by π minus five is the factored form of this quadratic.

We can of course confirm this by redistributing the parentheses. We then look for the critical values of this quadratic. And these are the values of π that make the expression equal to zero. We take each factor in turn and set it equal to zero. Giving π plus four equals zero and π minus five equals zero.

Each equation can be solved in a straightforward way. We solve the first equation by subtracting four from each side to give π equals negative four. And we solve the second by adding five to each side to give π equals five. And so we find that the critical values of this quadratic are negative four and five.

We can now use this information to help us sketch the graph of π¦ equals π squared minus π minus 20. We know that it is a quadratic. And as it has a positive leading coefficient β the coefficient of π squared is one β it will be a positive parabola. It has critical values of negative four and five. So these are the π₯-intercepts of the curve. And its π¦-intercept is the constant term of negative 20. Its turning point will be exactly halfway between the π₯-values of negative four and five. Thatβs 0.5. So itβs slightly to the right of the π¦-axis.

We can now use this graph to solve the quadratic inequality. Weβre looking for where our quadratic expression is less than zero. Thatβs below the π₯-axis. From our sketch, we can see that this occurs between the critical values of negative four and five. So we find that π squared minus π minus 20 is less than zero on the open interval negative four to five.

But this isnβt the solution to our overall inequality. Because remember, we said that π must also be nonzero. Otherwise, the quartic expression will give zero rather than something which is less than zero. We therefore need to subtract the single value zero from this open interval.

Another way to think about this would be as the union of two separate intervals. The open interval from negative four to zero and then the open interval from zero to five. So we can choose to express our answer in this way. The solution to the quartic inequality π to the fourth power minus π cubed minus 20π squared is less than zero. Is the union of the open interval negative four, zero and the open interval zero, five.