Knowing that the length of a rectangle is 𝑥 plus five and its area is two 𝑥 squared plus nine 𝑥 minus five, express the width of the rectangle algebraically.
This question is about a rectangle. We are given that the length of this rectangle is 𝑥 plus five. So let’s mark that on our diagram. We’re also told that the area is two 𝑥 squared plus nine 𝑥 minus five. So we write that in. And we have to find the width of this rectangle. How are we going to find the width of this rectangle from its length and area?
Well we know that for a rectangle, the area is equal to the length times the width. And by rearranging, we get that the width, which of course is what we want, is equal to the area divided by the length. And we have expressions for the area and length from the question. We substitute these expressions in to find that the width is two 𝑥 squared plus nine 𝑥 minus five over 𝑥 plus five.
We now have the width written as an algebraic fraction which we’ll see if we can simplify. There are two ways of simplifying this. We can either factor the numerator or we can perform a polynomial long division. We guess that we can factor the numerator into two 𝑥 plus something times 𝑥 plus something else.
Expanding this out and comparing coefficients, we see that 𝑎 plus two 𝑏 is nine and 𝑎𝑏 is negative five. Concentrating on the fact that 𝑎 times 𝑏 must be negative five, if 𝑎 and 𝑏 are integers, there are really only four possibilities. And going through each of these possibilities, we find that 𝑎 plus two 𝑏 is equal to nine as required when 𝑎 is negative one and 𝑏 is five.
So we substitute these values of 𝑎 and 𝑏 in. And so we see that two 𝑥 squared plus nine 𝑥 minus five is equal to two 𝑥 minus one times 𝑥 plus five. We could’ve used some other methods to factor this quadratic, for example, by using the quadratic formula to find the roots. And we could’ve guessed that one of the factors of this quadratic in the numerator would be 𝑥 plus five in order to cancel with the 𝑥 plus five in the denominator.
Speaking of which, we should use this factored form in the numerator. And we get that the width of the rectangle is simply two 𝑥 minus one. The method we used here to was to factor the numerator and then cancel. But we can actually solve this problem without factoring the numerator first. Instead, we can use polynomial long division. We first ask how many times 𝑥 goes into two 𝑥 squared. The answer is two 𝑥 times. And we then subtract two 𝑥 times our divisor 𝑥 plus five from our dividend.
But first, we need to expand this to make it easier to subtract. We have this subtraction laid out nicely in columns. Two 𝑥 squared minus two 𝑥 squared gives us zero 𝑥 squared, which we don’t bother writing down. Nine 𝑥 minus ten 𝑥 gives us minus 𝑥, which we write below in its own line, and negative five minus nothing is just negative five. So we end up with minus 𝑥 minus five.
And what we do now is to repeat the process. How many times does 𝑥 go into minus 𝑥? It goes negative one times. And now we have to subtract negative one times 𝑥 plus five. And what does this give us? Well subtracting this is the same as adding 𝑥 plus five. And performing this addition, we see that we get a remainder of zero.
That means that two 𝑥 squared plus nine 𝑥 minus five divided by 𝑥 plus five is two 𝑥 minus one exactly. There is no remainder. And interpreting this quotient in the context of the problem we had to solve, we see that the width of the rectangle, expressed algebraically, is two 𝑥 minus one.