Writing Polynomial Functions With Given Zeros - Revision history

Iwiseman: /* Example 3 */

2026-04-22T05:05:20Z

Example 3

← Older revision		Revision as of 05:05, 22 April 2026
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	So		So
	f(x)=30(x-2)(x + 3/4) (x²-10x+29)		f(x)=30(x-2)(x + 3/4) (x²-10x+29)
			But the interesting thing he did and the would make life easier was the x = -3/4 he times both sides by 4 to remove the fraction
			4x = -3 so 4x -3 = 0
			This made dealing with the problem a lot easier
			f(x) = a(x-2)(4x + 3)(x²-10x + 29)
			Plugging in 900 and 3
			900 = a(1)(12+3)(9-30 + 29)
			900 = a(1)(15)(8)
			15/2 = a
			This makes it far easier without the fractions inside the brackets
			f(x) = 15/2[(4x -5x -6)(x²-10x + 29)]
			= 15/2[4x⁴ - 40x³ + 116x² -5x³+50x²-145x -6x² + 60x - 174]
			= 15/2[4x⁴ - 45x³ +160x²--85x - 174]
			= 30x⁴ - 675/2x³ + 1200x² - 1275/2x - 1305

Iwiseman: /* Example 2 */

2026-04-22T04:45:08Z

Example 2

← Older revision		Revision as of 04:45, 22 April 2026
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	= 2(x³ +4x -3x² -12)		= 2(x³ +4x -3x² -12)
	= 2x³ - 6x² + 8x -24		= 2x³ - 6x² + 8x -24
			=Example 3=
			Gosh see if I improve
			n = 4 x=2,-3/4, 5-2i f(3) = 900
			So zero factors are
			(x-2)(x + 3/4)(x-5+2i)
			So
			f(x) = a(x-2)(x + 3/4)(x-5+2i)(x-5-2i)
			Doing
			(x-5-2i)(x-5+2i) = (x²-5x+2ix -5x+25-10i -2ix+10i-4i²)
			= x²-10x + 25 - 4i²
			So -4i² = 4 and therefore needed to explanation I could understand. And it is easy
			4i = -4 . (i.i)
			We know i² = -1 so
			4i = -4 . -1 = 4
			So back to the show
			= x²-10x + 29
			So
			f(x) = a(x-2)(x + 3/4)(x²-10x + 29)
			Plugging in 900 and 3
			900 = a(1)(3 + 3/4)(9-30 + 29)
			900 = a(3+3/4)(8)
			900 = a30
			30 = a
			So
			f(x)=30(x-2)(x + 3/4) (x²-10x+29)

Iwiseman at 03:35, 22 April 2026

2026-04-22T03:35:17Z

@@ Line 22: / Line 22: @@
 So the factors are
    (x - 3) and hmmmm
 i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>
-   = +/- <math>\sqrt{-2}</math>
+      = +/- <math>\sqrt{-2}</math>
-    = (x² - 4)
+      = (x² - 4)
-    = (x - 2)(x + 2)
+      = (x - 2)(x + 2)

Iwiseman: /* Example 2 */

2026-04-22T03:17:11Z

Example 2

← Older revision		Revision as of 03:17, 22 April 2026
Line 24:		Line 24:
	2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>		2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>
	= +/- <math>\sqrt{-2}</math>		= +/- <math>\sqrt{-2}</math>
	= (x² - 2)		= (x² - 4)
	= (x - 1)(x + 1)		= (x - 2)(x + 2)

Iwiseman: /* Example 2 */

2026-04-22T03:16:00Z

Example 2

← Older revision		Revision as of 03:16, 22 April 2026
Line 24:		Line 24:
	2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>		2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>
	= +/- <math>\sqrt{-2}</math>		= +/- <math>\sqrt{-2}</math>
			= (x² - 2)
			= (x - 1)(x + 1)

Iwiseman: /* Example 2 */

2026-04-22T03:14:24Z

Example 2

← Older revision		Revision as of 03:14, 22 April 2026
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	(x - 3) and hmmmm		(x - 3) and hmmmm
	2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>		2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>
			= +/- <math>\sqrt{-2}</math>

Iwiseman: Created page with "=Introduction= Hopefully something I can just do. I will try with the first question he had =Example 1= Write a polynomial with these characteristics n =2 (degree), x = 1, 2 (values with zero) and where f(3) = 6 So guessing we start with this is what it will look like f(3) = (x-1)(x-2) = 6 = (x-1)(x-2) - 6 = 0 = x² -3x -6 And wrong. Like being wrong because I hopefully will remember this and not make the same mistake again. So where did I go wrong. We need..."

2026-04-22T03:13:48Z

Created page with "=Introduction= Hopefully something I can just do. I will try with the first question he had =Example 1= Write a polynomial with these characteristics n =2 (degree), x = 1, 2 (values with zero) and where f(3) = 6 So guessing we start with this is what it will look like f(3) = (x-1)(x-2) = 6 = (x-1)(x-2) - 6 = 0 = x² -3x -6 And wrong. Like being wrong because I hopefully will remember this and not make the same mistake again. So where did I go wrong. We need..."

New page

=Introduction=
Hopefully something I can just do. I will try with the first question he had
=Example 1=
Write a polynomial with these characteristics
n =2 (degree), x = 1, 2 (values with zero) and where f(3) = 6
So guessing we start with this is what it will look like
f(3) = (x-1)(x-2) = 6
= (x-1)(x-2) - 6 = 0
= x² -3x -6
And wrong. Like being wrong because I hopefully will remember this and not make the same mistake again. So where did I go wrong. We need another step to find the leading coefficient a so we write
f(x) = a(x-1)(x-2)
So we know when x = 3 the function = 6
6 = a(3-1)(3-2)
6 = 2a
3 = a
So
f(x) = 3(x-1)(x-2)
= 3[x² -3x +2]
= 3x² -9x + 6
=Example 2=
n = 3 x=3,2i f(4) = 40
So the factors are
(x - 3) and hmmmm
2i = +/- <math>\sqrt{2}</math> * <math>\sqrt{-1}</math>