Show that the real zeros of each polynomial function satisfy the ...

Question

Show that the real zeros of each polynomial function satisfy the given conditions. See Example 6. ƒ(x)=x^5-3x^3+x+2; no real zero less than -3

Accepted Answer

Mhm Hey, everyone in this problem for the following polynomial function, we're asked to determine whether those real zeros satisfies the given condition or not. We're given the function F of X is equal to three X to the exponent five plus five X cubed minus seven X squared plus two X plus nine. We're told that there is no real zero less than negative five. Now we have two answer choices here. Option A yes or option B no. So we need to figure out whether this statement is true, whether the real zeroes satisfy this condition that we've been given. Now this statement no real zero less than negative five. So that's putting a lower bound on the real zeros. Now our call we have something called the lower bound theorem. Yeah. And sometimes it's called the bounded theorem. So you may have seen either term. Now this is a really neat zero. OK. It tells us that if we take our function F of X, we use synthetic division and divide it by what we're thinking is a lower bound. So in this case negative five, and we look at that bottom row in our synthetic division table, if the signs of those values alternate, positive, negative, positive negative, then that is a lower bound on our real zeros. So what we're gonna do is we're gonna set up a synthetic division table. OK? It's gonna have two ropes on the left hand side, we're gonna have the value that we're testing as a lower bound, which is negative five in the top of our table. We're gonna have the coefficients corresponding with the terms in our function F of X starting on the left hand side, the coefficient corresponding with the highest degree term all the way down to our constant term on the right hand side. So the highest exponent we have is five X to the exponent five. And so the coefficient corresponding to that is three. So we have three. Now the next highest coefficient will be four. OK? Er Sorry, not next highest coefficient, next highest exponent will be four and we don't have an X to the four term. So the coefficient is actually zero and it's really important that you include the zero and you can't skip it and go just to the X Q term. Now moving to the X cube term coefficient of five to the X squared term coefficient of negative seven to the X term coefficient of two and the constant term A coefficient of nine. OK. So that first row from left to right, 305, negative 729. Now we're gonna get started with the actual division. So we're gonna start on the left and we're gonna bring this first column, first row value of three down. And underneath our table, we have a value of three underneath our table. When we have a value under our table, we multiply it to the value on the left of our table, three multiplied by negative five gives us negative 15. We're gonna put that in the next column over and in the second row. So now we have a completed column. We're gonna add within that call zero plus negative 15 gives us negative 15. We write that down below. We have a new number down below. So we multiply it to the number on the left hand side, negative 15, multiplied by five gives us 75. We write it in the next column over in the second row. And now we have another completed column and we're gonna add within that column five plus 75 gives us 80. We write it down below a new number down below. We multiply to the number on the left 80 multiplied by negative five gives us negative 400. We write it in the column over second row and yet we have another completed column. We'll add within that column, negative seven plus negative 400 gives us negative 407. Write it down below a new number down below. We multiply it to the number on the left hand side negative multiplied by negative five. Gives us 2035. We write it in the next column over second row. We have another completed column and we add two plus 2035 gives us 2037. Write it down below new number, down below multiplies to that number on the left multiplied by negative five, gives us negative 10,185. Again, we write it in the next column over in the second row, we have our final column complete and we're gonna add within that column nine plus negative 10,185 gives us negative 10,176. Now, if we look at this bottom row in our synthetic division table, now that we're done, we have three negative 15, 80 negative 407 negative one or negative 10,176. OK. So in this bottom row, the signs alternate, hey, it goes positive, negative, positive, negative, positive negative. So that tells us by our lower bound theorem that negative five is a lower bound on our zeroes are real zeroes. If it's a lower bound on our real zeros, what that means is that there are no real zeros less than negative five. And so this statement is true. We have answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with College Algebra?

Show that the real zeros of each polynomial function satisfy the given conditions. See Example 6. ƒ(x)=x^5-3x^3+x+2; no real zero less than -3

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