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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
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