Video: Simplifying Polynomials by Division to Express the Width of a Rectangle

What is the width of a rectangle whose area is (4π‘₯Β³ βˆ’ 32π‘₯Β² + 6π‘₯⁴) cmΒ² and whose length is (8π‘₯ + 3π‘₯Β²) cm?

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Video Transcript

What is the width of a rectangle whose area is four π‘₯ cubed minus 32π‘₯ squared plus six π‘₯ to the fourth centimeters squared and whose length is eight π‘₯ plus three π‘₯ squared centimeters?

Here’s our rectangle whose area is four π‘₯ cubed minus 32π‘₯ squared plus six π‘₯ to the fourth centimeters squared and whose length measures eight π‘₯ plus three π‘₯ squared centimeters. And we’re trying to find the width. We know that finding the area of the rectangle is found by multiplying the length times the width. And we can substitute values in for what we know.

Four π‘₯ cubed minus 32π‘₯ squared plus six π‘₯ to the fourth equals eight π‘₯ plus three π‘₯ squared times the width. To find 𝑀, we can divide by eight π‘₯ plus three π‘₯ squared. And if we do that to one side, we have to do it to the other side. On the right side, everything cancels except the 𝑀. So we’ll have 𝑀 equals four π‘₯ cubed minus 32π‘₯ squared plus six π‘₯ to the fourth over eight π‘₯ plus three π‘₯ squared. What we’re saying here is the width equals the area divided by the length.

There are a few ways to solve this. One of the ways would be to factor both the numerator and the denominator before we try to divide. We’ll give ourselves some space to factor the numerator, four π‘₯ cubed minus 32π‘₯ squared plus six π‘₯ to the fourth. The exponents of this expression are not in descending order.

The first thing we want to do is rearrange them so that the degree of each exponent is descending. We’ll start with six π‘₯ to the fourth plus four π‘₯ cubed minus 32π‘₯ squared. Is there any factors that each of these terms have in common? They’re all even numbers, which means they all share a factor of two, and they all have at least π‘₯ squared. We can factor out the term two π‘₯ squared. When we take out two π‘₯ squared, six π‘₯ to the fourth becomes three π‘₯ squared, four π‘₯ cubed becomes two π‘₯, and negative 32π‘₯ squared becomes negative 16. We’ll bring down our two π‘₯ squared. And then we’ll consider if three π‘₯ squared plus two π‘₯ minus 16 can be factored any further.

Three π‘₯ squared can be factored into three π‘₯ and π‘₯. We need two values that multiply together to equal negative 16. One and negative 16, or negative one positive 16, two and negative eight, negative two and positive eight, or positive four and negative four. The key here is that our middle term must be equal to positive two and that one of these values has to be multiplied by three. One and negative 16 is going to be too big. Let’s try two and eight. If we put negative eight and positive two, our middle term becomes six π‘₯ minus eight π‘₯, which is negative two π‘₯.

We’re nearly there. Because we have negative two, we realize that we’re looking for a negative six π‘₯ and a positive eight π‘₯. The factorized form of the numerator is then two π‘₯ squared times three π‘₯ plus eight times π‘₯ minus two. We now need to factor the denominator. What factors do eight π‘₯ and three π‘₯ squared have in common? They share an π‘₯. And if we take that π‘₯ factor out, we’ll be left with eight plus three π‘₯. And again, we like our degrees in descending order. So let’s switch that around and write three π‘₯ plus eight.

It’s time to substitute these values back into our fraction. Two π‘₯ squared times three π‘₯ plus eight times π‘₯ minus two over π‘₯ times three π‘₯ plus eight. We have a three π‘₯ plus eight in the numerator and the denominator, so they cancel out. And we have an π‘₯ term in the numerator and the denominator. The π‘₯ in the denominator cancels out, and the π‘₯ squared in the numerator becomes π‘₯. There’s nothing remaining in the denominator.

So we have a statement that says the width equals two π‘₯ times π‘₯ minus two, and we wanna distribute this two π‘₯. Two π‘₯ times π‘₯ equals two π‘₯ squared. Two π‘₯ times negative two equals negative four π‘₯. The width equals two π‘₯ squared minus four π‘₯ centimeters.

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