In Exercises 9–16, a) List all possible rational zeros. b) Use sy...

Question

In Exercises 9–16, a) List all possible rational zeros. b) Use synthetic division to test the possible rational zeros and find an actual zero. c) Use the quotient from part (b) to find the remaining zeros of the polynomial function. f(x)=2x^3−3x^2−11x+6

Accepted Answer

Hey everyone here we are asked to this the possible rational zeros and find an actual zero of the function F of X is equal to two, X cubed plus X squared minus X plus 20. And here to start finding the possible rational zeros. We first have to find the factors of the constant as well as the factors of the leading coefficient. So starting with the constant which is 12. Well 20 has factors of 1245, 10 and 20. And each of these values can be, each of these factors can be either positive or negative, so they each get a plus or minus sign in front of them and now we can move on to finding the factors of our leading coefficient which is +22 has factors of just one and two. And again both of these values get a plus or minus sign. Now we just have to recall the rational zeros theorem which tells us that we can take the factors of the constant value and divide it by the factors of our leading coefficient. So doing this, we have first dividing by one. We have one divided by one giving us plus or minus one, then two divided by one giving us plus or minus +24 divided by one which is plus or minus four, then we have plus or minus five plus or minus 10 and plus or minus 20. And now we can move on to dividing each value by two. So we have one divided by two giving us plus or minus one half. We then have two divided by two which would give us plus or minus one. But we already have this so we don't need to rewrite it. We then have four divided by two giving us plus or minus two. Which again we already have, we then have plus or minus five divided by plus or minus two. Giving us plus or minus five halves. Then we have 10 divided by two giving us plus or minus five. Which again we already have, we then have plus or minus 20 divided by plus or minus two giving us plus or minus 10, which again we already have. Therefore we're done finding our list of possible rational zeros and we can move on to using synthetic division to find the actual zero. So for synthetic division, if you recall on the right hand side we have our coefficients from our given equation. So we have 21 negative 41 positive 20. And on the left hand side we choose one of our possible rational zeros. And so I'm going to choose positive four. And now just solving this as you normally would with synthetic division, we first bring down the first coefficient which is two and don't make any changes to it. And then we bring it up and we multiply it by our left hand side value which is four. So we have two times four which gives us eight. We then added to the other coefficient in the column. So we have eight plus one which gives us nine. We then repeat this process. I mean take nine times four giving us positive 36 36 plus negative 41 gives us negative five. We then take negative five and multiply it by four. So we get negative 20 negative 20 plus positive 20 gives us zero. So we see that we have a remainder of zero and therefore we know that this positive four here is an actual zero and so now we can find the rest of the remaining zeros by setting our original equation equal to zero. So we have two X cubed plus X squared minus X plus 20 is equal to zero. And now we can factor out this actual zero we found. So we have the quantity of x minus four times the quantity of two, X squared plus nine, X minus five is equal to zero. And we can continue to factor this. And so we get the quantity of x minus four times the quantity of two x minus one times the quantity of X plus five. And now from here we can set each set of parentheses equal to zero to find all of our remaining zeros. So starting with x minus four, what we already found this earlier but we see again that X is equal to positive four. So we know that this is a zero and now we can move on to two, X minus one is equal to zero, adding one to both sides. We have two. X is equal to one, dividing both sides by two. We see that X is equal to one half, and finally with X plus five is equal to zero. We see that X is equal to negative five. So in summary all of our actual zeros here we have X is equal to negative +51 half and positive four. And now we have found the actual zeros as well as the list of possible rational zeros from earlier. Therefore we are left with answer choice B. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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In Exercises 9–16, a) List all possible rational zeros. b) Use synthetic division to test the possible rational zeros and find an actual zero. c) Use the quotient from part (b) to find the remaining zeros of the polynomial function. f(x)=2x^3−3x^2−11x+6