Solve each polynomial equation in Exercises 1–10 by factoring ...

Question

Solve each polynomial equation in Exercises 1–10 by factoring and then using the zero-product principle. $4 y^{3} - 2 = y - 8 y^{2}$

Accepted Answer

Hi. So this problem says the use factoring in the zero product principle to solve the given polynomial equation, we have four beats in the third plus eight beats in the second equals nine B plus 18. So what I first want to do is get everything equal to zero. So I'm gonna move myself on the right side of the equal sign to the left. I get four weeks of the third plus eight B squared. I'm gonna subtract nine B and 18 from both cents. So I would get negative nine B minus 18 equals zero. Now we're gonna try and factor out this polynomial. So if you look here, let's look at four BQ plus 80 square first. Is there anything in common? We can factor out of that polynomial? Well let's see. We have a four and an eight. So let's say we can take out a four. We also have to be cute. And then B squared. Let's say we take out B squared from that. We do that. We're left with four B squared on the outside. We took out which we have B plus two left on the inside. And I can do the same for the nine. B -18. We factor this we should have a term. That's the same as B plus two. So let's say we factor out a negative nine. Next nine we take out negative nine we're left with the B. They're taking a negative nine out of a negative 18 gives us positive too. Just be equal to zero. So now we can go ahead and combine our terms. So we can combine our four B squared are negative nine since they are outside the parentheses and we can go and combine our B plus two. S until one, B plus two. Now we have four B squared minus nine times B plus two is equal to zero. The next thing we can do is factor out this four, B squared minus nine. We can make use of the difference of two squares there. And so if we have a squared minus B squared, the factors are A plus B. Hey you mind speed. We have four B squared months. None. And so are A in this case would be to be And RB in this case would be three. So four B square minus nine. That turns into two, B plus three And to be -3. We can rewrite that in our main polynomial, We have to be plus three to be minus three and B plus two. All equal zero. Now we can go ahead and set every one of our factors equal to zero. So let's say we have to be plus three equals zero to be minus three equals zero and B plus two equals zero. Let's go ahead. And software be for each of these have minus three here, which means we get to b equals negative three. Finally dividing by two gives us B equals negative three halves. Same thing here we add three to both sides get to B equals three by two. B equals three. EFS. Finally, we have B plus two equals zero. Subtract two from both sides. We get B equals negative two. So the answer is for our B our vehicles. Think of two vehicles, three halves and B equals negative three halves, which corresponds to answer D. Alright, I hope to help you solve the problem. Thank you for watching. Goodbye.

Solve each polynomial equation in Exercises 1–10 by factoring and then using the zero-product principle. $4 y^{3} - 2 = y - 8 y^{2}$

Key Concepts

Polynomial Equations

Factoring Polynomials

Zero-Product Principle

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Solve each polynomial equation in Exercises 1–10 by factoring and then using the zero-product principle. 4y3−2=y−8y24y^3 - 2 = y - 8y^2

Key Concepts

Polynomial Equations

Factoring Polynomials

Zero-Product Principle

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Solve each polynomial equation in Exercises 1–10 by factoring and then using the zero-product principle. $4 y^{3} - 2 = y - 8 y^{2}$