The Remainder Theorem - Revision history

Iwiseman: /* Example 2 */

2026-04-21T03:22:17Z

Example 2

← Older revision		Revision as of 03:22, 21 April 2026
Line 34:		Line 34:
	= 967		= 967

	5 \| 3 - 7 0 - 9 + 12		5 \| 3 - 7 0 - 9 + 12
	\| ↓ 15 40 200 + 955		\| ↓ 15 40 200 + 955
	-----------------------		------------------------
	3 8 40 191 = 967		3 8 40 191 = 967
			3x³+8x²+40x+191 = 967
	So		So

Iwiseman: /* Example 1 */

2026-04-21T03:17:35Z

Example 1

← Older revision		Revision as of 03:17, 21 April 2026
Line 23:		Line 23:
	\| ↓ + 8 + 12 + 72		\| ↓ + 8 + 12 + 72
	-----------------		-----------------
	~~2x³~~+3x + 18 = 60		2x²+3x + 18 = 60

	=Example 2=		=Example 2=

Iwiseman at 02:01, 21 April 2026

2026-04-21T02:01:54Z

← Older revision		Revision as of 02:01, 21 April 2026
Line 41:		Line 41:

	*the divisor is x − 5, so a = 5		*the divisor is x − 5, so a = 5
	~~•the~~ polynomial is f(x) = 3x⁴ − 7x³ + 0x² − 9x + 12		*the polynomial is f(x) = 3x⁴ − 7x³ + 0x² − 9x + 12
	~~•the~~ computed f(5) = 967		*the computed f(5) = 967
	~~•the~~ synthetic division remainder = 967		*the synthetic division remainder = 967

	So the theorem becomes:<br>		So the theorem becomes:<br>

Iwiseman: Created page with "=Introduction= No idea where this is going but here goes. They start with an example. Asked the robot what is was and it said If you divide a polynomial f(x) by (x-a) the '''remainder''' is simply f(a) =Example 1= In this we appear to use the approach of substitution and synthetic division f(x) = 2x³ - 5x² + 6x -12 We are asked to workout when f(4). Using substitution we get f(4) = 2(4)³ - 5(4)² + 6(4) -12 = 128 - 80 + 24 - 12 f(4) = 60 They went of to..."

2026-04-21T02:01:37Z

Created page with "=Introduction= No idea where this is going but here goes. They start with an example. Asked the robot what is was and it said If you divide a polynomial f(x) by (x-a) the '''remainder''' is simply f(a) =Example 1= In this we appear to use the approach of substitution and synthetic division f(x) = 2x³ - 5x² + 6x -12 We are asked to workout when f(4). Using substitution we get f(4) = 2(4)³ - 5(4)² + 6(4) -12 = 128 - 80 + 24 - 12 f(4) = 60 They went of to..."

New page

=Introduction=
No idea where this is going but here goes. They start with an example. Asked the robot what is was and it said
If you divide a polynomial f(x) by (x-a) the '''remainder''' is simply f(a)

=Example 1=
In this we appear to use the approach of substitution and synthetic division

f(x) = 2x³ - 5x² + 6x -12
We are asked to workout when f(4). Using substitution we get
f(4) = 2(4)³ - 5(4)² + 6(4) -12
= 128 - 80 + 24 - 12
f(4) = 60

They went of to use synthetic division so write down the coefficient and bring down the 2

4 | 2 - 5 + 6 - 12
| ↓
-----------------
2

Continue Multiplying and adding
4 | 2 - 5 + 6 - 12
| ↓ + 8 + 12 + 72
-----------------
2x³+3x + 18 = 60

=Example 2=
Here we go again
f(x) = 3x⁴ - 7x³ + 0x² - 9x +12 where f(5)
= 3(5)⁴ -7(5)³ + 0 -9(5) + 12
= 3(625) -7(125) - 45 + 12
= 1875 - 875 -33
= 1000 - 33
= 967

5 | 3 - 7 0 - 9 + 12
| ↓ 15 40 200 + 955
-----------------------
3 8 40 191 = 967
So

*the divisor is x − 5, so a = 5
•the polynomial is f(x) = 3x⁴ − 7x³ + 0x² − 9x + 12
•the computed f(5) = 967
•the synthetic division remainder = 967

So the theorem becomes:<br>
If you divide the polynomial 3x⁴ − 7x³ + 0x² − 9x + 12 by (x-5),<br>
the remainder is simply f(5)=967.<br>

And that matches your synthetic division exactly.