Video: US-SAT03S3-Q13-845191629734

In the equation, (2π‘šΒ² + 3π‘š βˆ’ 9)/(π‘šΒ² + 7π‘š + 12) = (π‘šΒ² βˆ’ 6π‘š + 5)/(π‘šΒ² βˆ’ π‘š βˆ’ 20). What is the value of π‘š?


Video Transcript

In the equation, two π‘š squared plus three π‘š minus nine over π‘š squared plus seven π‘š plus 12 equals π‘š squared minus six π‘š plus five over π‘š squared minus π‘š minus 20. What is the value of π‘š?

In this question, we have four different polynomials. The best way to solve this is to factor each of these polynomials and see if there’s anything we can cancel out before we try to solve for π‘š. First we’ll factor the polynomial on the top left. Since we have a first term of two π‘š squared, we’ll break that up into two π‘š and π‘š. And then we need two values that multiply together to equal negative nine and that add together to equal positive three.

I know that three times three equals nine. Since we have negative nine, we need a positive three and a negative three. And two π‘š times three equals positive six π‘š. Negative three times π‘š equals negative three π‘š. And so we get our positive three π‘š we need. And now, we factor the denominator. We break up the π‘š squared as π‘š times π‘š. We need values that multiply together to equal 12 and add together to equal seven. Four times three multiplies together to equal 12 and adds together to equal positive seven.

From there, we factor the numerator on the second side, π‘š squared. We break that apart to be π‘š and π‘š. We’re looking for values that multiply together to equal five and added together equals negative six. Five only has two factors, five and one. We need these values to add together to equal negative six. That means both the five and the one should be negative. Negative five times negative one still equals positive five. We need to factor our final polynomial. π‘š squared will factor as π‘š and π‘š. Our terms need to multiply together to equal negative 20 and add together to equal negative π‘š. Or in this case, we’re looking for negative one.

I know that five times four equals 20 and that negative five plus four would equal negative one. Positive four times negative five equals negative 20. And so we have our final two terms. And at this point, we start to see some things that cancel out. We have an π‘š plus three in the numerator and the denominator on the left. And we have π‘š minus five in the numerator and the denominator on the right.

We simplify this to say two π‘š minus three over π‘š plus four equals π‘š minus one over π‘š plus four. If we multiplied both sides of the equation by π‘š plus four, the denominators would cancel out, so that we have two π‘š minus three equals π‘š minus one. To get the π‘šs on the same side, we subtract π‘š from both sides. π‘š minus three equals negative one. And so we add three to both sides. And we find that π‘š equals two.

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