Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
