G. "I can use the remainder and factor theorems to help factor polynomial equations of degree 3 or 4." 1. x²-x-x-10 by *-2 2. *²¹-28x-41 3. 2x²-3x²-8x-3 by x²-3x-7 x-2x - 5x²-x-10 x'-2₁² ↓ ↓ -3x²-x + -3x²+6x -7x-10 -7x+14 -24 P(2)= Now, for P(x) try P(x)=x²-5x²-x-10 x = 2 1-4x-12 x+ 4x + 0x - 28-41 x² + 4x² by - 4x² -28.r -4x²-16.x -12x-41 - 12.x - 48 7 x+4 Now, for P(x) try = -4 P(x) = x³ - 28x-41 P(-4)= x-2x-3 2x+12x²-3x²-8x-3 2x¹+x² ↓ ↓ -4.² -8.x -4.x² - 2x -6x -3 -6.x-3 0 Now, for P(x) try*=-1 P(x) = 2x¹³ - 3x²-8x-3 P(-) =

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter8: Polynomials
Section8.6: Factoring Quadratic Trinomials
Problem 61PFA
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Question
L.G. "I can use the remainder and factor theorems to help factor polynomial equations of degree 3 or 4."
1. x² -5x²-x-10 by *-2
2. *¹-28x-41 by *+4
3. 2x -3x²-8x-3 by
x²-3x-7
x-2x²¹ -5x²-x-10
x'-21² ↓
-3x²-x
-3x²+6x
↓
P(2)=
-7x-10
-7.x+14
- 24
Now, for P(x) try * = 2
P(x)=x²-5x²-x-10
-4.x-12
x+ 4x + 0-28-41
X
² + 4x²
2
- 4x² -28.r ↓
-4x²-16.x
P(-4)=
- 12.x-41
- 12x - 48
7
Now, for P(x) try
P(x)=x²-28x-41
x = -4
-2x-3
2x+1|2x - 3x²-8r-3
2.x²+x² ↓ ↓
-4.x²-8.x
-4.x² - 2x +
- 6x -3
-6.x-3
0
Now, for P(x) try* = -1
P(x) = 2x¹ - 3x²-8x-3
P(-) =
2x+1
Transcribed Image Text:L.G. "I can use the remainder and factor theorems to help factor polynomial equations of degree 3 or 4." 1. x² -5x²-x-10 by *-2 2. *¹-28x-41 by *+4 3. 2x -3x²-8x-3 by x²-3x-7 x-2x²¹ -5x²-x-10 x'-21² ↓ -3x²-x -3x²+6x ↓ P(2)= -7x-10 -7.x+14 - 24 Now, for P(x) try * = 2 P(x)=x²-5x²-x-10 -4.x-12 x+ 4x + 0-28-41 X ² + 4x² 2 - 4x² -28.r ↓ -4x²-16.x P(-4)= - 12.x-41 - 12x - 48 7 Now, for P(x) try P(x)=x²-28x-41 x = -4 -2x-3 2x+1|2x - 3x²-8r-3 2.x²+x² ↓ ↓ -4.x²-8.x -4.x² - 2x + - 6x -3 -6.x-3 0 Now, for P(x) try* = -1 P(x) = 2x¹ - 3x²-8x-3 P(-) = 2x+1
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