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

Question

Show that the real zeros of each polynomial function satisfy the given conditions. ƒ(x)=x⁵-3x³+x+2; no real zero greater than 2

Accepted Answer

Hey, everyone in this problem for the following polynomial function determine whether the real zero satisfies the given condition or not. Now, the function we're given is F of X is equal to four X to the exponent five minus three X to the exponent four plus five X cubed minus seven X squared plus 12, X minus 11. And the condition we're given is that no real zero is greater than four. OK. And we're given two options, answer a yes or answer B no. So this no real greater, no real zero, greater than four. OK. Means that we have this upper bound on our real zeros. And what we wanna do is we want to consider this really neat theorem called the upper bound zero. And it's sometimes also referred to as the bounded theorem. Now, what this theorem tell us is that if we take a function F of X and we divide it by the value that we're looking at for our bound. So we're gonna divide it by four using synthetic division. And we look at that last row in our synthetic division table. If all of the values in that row are non negative. OK. Then that number we divided our function by is an upper bound on the real zeros. OK. So we're gonna do that. We're gonna draw our synthetic division table and we're gonna take our function F of X and divide it by four. So the value on the left-hand side of our table is going to be that number four. OK. Our table has two rows in the first row. We're gonna fill in the coefficients of our function F of X. We're gonna start on the left hand side with the coefficient corresponding to the highest exponent term and work our way to the right to the constant term. So X to the exponent five, that's our leading order term has a coefficient of four. And we have our X to the exponent four term with a coefficient of negative three or X cube term with a coefficient of five are X squared term with a coefficient of negative seven, our X term with a coefficient of 12 and finally a constant term of negative. So the top row of our table from left to right is four negative 35, negative 7, 12 and negative 11. And now we're gonna perform the synthetic division. So starting on the left hand side, we bring that number four down. OK, from our top row and we write it underneath our table. When we have a value underneath our table, it's gonna multiply to the value on the left that's also outside of our table. Four, multiplied by four gives us 16. We're gonna write it in the next column over in the second row of the table. And now we add within that call. So we have negative three plus 16, which gives us 13. We write it down below. Now we have this new value of 13 below our table. We multiply it to the value on the left 13 multiplied by four, gives us 52. We write it in the next column over in the second row. And now we add in the column five plus gives us 57. We write it down below you multiply it to the value on the left. 57 multiplied by four, gives us 228 plus negative seven K. Adding in our column gives us 221. We write it down below. We take our value underneath, we multiply it to the value on the left 2 21 multiplied by four. Gives us 884. We write it in the next column over the second row and we add within that column, 12 plus 884 gives us 896. OK? We write it down below that new value. Down below it is gonna multiply the value on the left multiplied by four, gives us 3584 we write it in the next column over. And now we are in our last column, we can add up that last column. We have negative 11 plus and we get 3573. Now, with this upper bound theorem, what we wanna do is look at this bottom row in our synthetic division table. And when we look at that bottom row, we can see that all of these values turn on N OK. They're all positive values in this case, which tells us that or is an upper bound on the real zeroes. Now, if four is an upper bound on the real zeros, that means that there are no real zeros greater than four. So that statement or that condition we were given is true. OK? We satisfy that statement. And so the correct answer is option A yes. Thanks everyone for watching. I hope this video helped see you in the next one.

Show that the real zeros of each polynomial function satisfy the given conditions. ƒ(x)=x⁵-3x³+x+2; no real zero greater than 2

Key Concepts

Real Zeros of Polynomial Functions

Evaluating Polynomial Values to Bound Zeros

Intermediate Value Theorem

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Show that the real zeros of each polynomial function satisfy the given conditions. ƒ(x)=x5-3x3+x+2; no real zero greater than 2

Key Concepts

Real Zeros of Polynomial Functions

Evaluating Polynomial Values to Bound Zeros

Intermediate Value Theorem

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Show that the real zeros of each polynomial function satisfy the given conditions. ƒ(x)=x⁵-3x³+x+2; no real zero greater than 2