00:01
Okay, i am going to use synthetic division to find each of these function values because we know according to the remainder theorem that the remainder from synthetic division is equal to the function value.
00:12
So for part a, for the synthetic division, we put one outside the box, and then we have 2x cubed, 0x squared, negative 7x, and 3.
00:23
We bring down the 2, multiply it by 1, write it in the next space, and add the column.
00:29
Multiply that by 1, write it in the next space and add the column.
00:33
Multiply that by 1, write it in the next space and add the column.
00:37
So the remainder is negative 2.
00:39
So that means that f of 1 is negative 2.
00:45
For part b, we do the same thing to find f of negative 2.
00:51
Negative 2 goes outside the box.
00:54
2, 0, negative 7, and 3 go in a row.
00:57
Bring down the first number.
00:59
Multiply it by negative 2, write it down and add.
01:02
Multiply that by negative 2, write it down and add.
01:05
Multiply that by negative 2, write it down and add.
01:13
Okay, so the remainder is 1, so that means f of negative 2 is 1.
01:28
Okay, let's go again with part c, and we're finding f of 1 half.
01:34
So 1 half is outside the box.
01:37
We have 2, 0, negative 7, and 3.
01:41
Bring down the 2, multiply it by 1 half, it gives us 1, write it down and add.
01:46
Multiply that by 1ā2, it gives us 1ā2, write it down and add.
01:50
Okay, so now we have negative 6 .5, and i'm going to write that as negative 13 halves.
01:58
Multiply that by 1 half, and we get negative 13 fourths.
02:03
And when we add that to 3, think of 3 as 12 4ths.
02:07
So we're adding negative 13 4ths to 12 4ths, so we have negative 1 4th...