Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Comparing x-1 with x-a then x=1
1^3+p*1^2-1-2=0
1+2p-3=0
2p=3-1
P=2/2
P=1
:.p=1
1^3+p*1^2-1-2=0
1+2p-3=0
2p=3-1
P=2/2
P=1
:.p=1
The polynomial [tex]x^3+ px^{2}-x-2[/tex] contains x - 1 as a factor.
The value of p = 2.
How to find the value of p?
Given:
The polynomial [tex]x^3+ px^{2}-x-2[/tex] contains x - 1 as a factor.
Comparing x - 1 with x - a then
Substitute the value of x = 1, then we get
[tex]$1^3+p*1^2-1-2 = 0[/tex]
[tex]$1^{3}+p \cdot 1^{2}-1-2=0$[/tex]
Apply rule [tex]$1^{a}=1$[/tex]
[tex]${data-answer}amp;1^{3}=1,1^{2}=1 \\[/tex]
[tex]${data-answer}amp;1+1 \cdot p-1-2=0[/tex]
Multiply: p-1 = p
1+p -1-2=0
Group like terms
p+1-1-2 = 0
Add/Subtract the numbers: 1-1-2 = -2
p-2 = 0
Add 2 to both sides
p-2+2 = 0+2
Simplify
p = 2
Therefore, the value of p = 2.
To learn more about polynomial function
https://brainly.com/question/11434122
#SPJ2
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
