Answer each question. By what expression should we multiply each ...

Question

Answer each question. By what expression should we multiply each side of (3x - 2)/(x + 4)(3x^2 + 1) = A/(x + 4) + (Bx + C)/(3x^2 + 1) so that there are no fractions in the equation?

Accepted Answer

Welcome back. I am so glad you're here. We are asked what should be multiplied to each side of them. We're given an equation. I'll read that in a minute so that the fraction form of the equation is removed. Here's our equation. On the left hand side of the equal sign, we have one fraction in the numerator. We have eight Z minus one that's divided by the quantity of Z plus five multiplied by the quantity of two Z squared plus seven. Now that's equal to, on the right hand side, we have two fractions. The first in the numerator is capital A that's divided by Z plus five. Then plus the second fraction in the numerator, we have BZ plus C which is divided by, in the denominator two Z squared plus seven. Our answer choices are answer choice A, the quantity of Z plus five multiplied by the quantity of two Z squared plus seven. Answer choice B, the quantity of Z plus seven multiplied by the quantity of two Z squared plus five. Answer choice C, the quantity of Z plus two multiplied by the quantity of five Z squared plus seven. And to answer choice D, the quantity of two Z squared plus seven. All right. So we need to make this, all of these fractions go away. And in order to do that, we're going to multiply everything by the factors that occur through all of our denominators. So looking on the right hand side, the first fraction has a denominator with the quantity of Z plus five. And that second fraction has a denominator of two Z squared plus seven. That's not the same as the first one. So we're going to multiply it by the two Z squared plus seven as well. And looking at the left hand side of the equation, that's exactly the denominator that they have there. So multiplying everything by the quantity of Z plus five, multiplied by the quantity of two Z squared plus seven will get rid of the fractions. And even though you know the answer, now I'll show you how that works. So we'll have the numerator from the left hand side, we have eight Z minus one. Now that's being multiplied by, we'll keep everything as factors. So the quantity of Z plus five multiplied by the quantity of two Z squared plus seven. And that's all divided by the quantity of Z plus five multiplied by the quantity of two Z squared plus seven that equals on the right, we have the capital A now being multiplied by the quantity of Z plus five multiplied by the quantity of two Z squared plus seven and that's divided by Z plus five. That fraction is plus. Now we have the numerator of the second fraction. On the right hand side, we have BZ plus C which is now being multiplied by the quantity of Z plus five and multiplied by the quantity of two Z squared plus seven all divided by our denominator. Oops switch colors which is two Z squared plus seven. Sorry if my twos and Zs look alike. So now is the beautiful art of canceling things out. On the left hand side. That fraction, the Z plus five factors cancel out and the two Z squared plus seven factors cancel out on the right hand side. The first fraction, the Z plus five factors cancel out and the second fraction, the two Z squared plus seven factors cancel out and look at that. Our denominators are gone, our fractions are gone and that is why that answer choice works. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

Answer each question. By what expression should we multiply each side of (3x - 2)/(x + 4)(3x^2 + 1) = A/(x + 4) + (Bx + C)/(3x^2 + 1) so that there are no fractions in the equation?

Key Concepts

Common Denominator

Rational Expressions

Algebraic Manipulation

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