Answer:
1) [tex]\boxed{p(x) = x^3-x^2+x-1}[/tex]
2) [tex]\boxed{p(x) = x^2+x-2}[/tex]
3) [tex]\boxed{p(x) =- 2x^2+2x+4}[/tex]
4) [tex]\boxed{p(x) = 2x^2+x-4}[/tex]
Step-by-step explanation:
Part (1)
[tex]p(x) = x^3-x^2+x-1[/tex]
As we have to determine it by ourselves, this is the polynomial having a degree of 3. p(x) with a degree of 3 means that the highest degree/exponent of x should be 3.
Part (2)
[tex]p(x) = x^2+x-2[/tex]
This can be the polynomial having the factor x-1 because if we put:
x - 1 = 0 => x = 1 in the above polynomial, it gives us a result of zero which shows us that (x-1) "is" a factor of the polynomial.
Part (3)
[tex]p(x) = -2x^2+2x+4[/tex]
This can be the polynomial for which p(0) = 4 and p(-1) = 0
Let's check:
[tex]p(0) =- 2(0)^2+2(0)+4\\p(0) = 0 + 0+4\\p(0) = 4[/tex]
[tex]p(-1)= -2(-1)^2+2(-1)+4\\p(-1) = -2(1)-2+4\\p(-1) = -2-2+4\\p(-1) = 0[/tex]
So, this is the required polynomial determined by "myself".
Part (4):
[tex]p(x) = 2x^2+x-4[/tex]
This is the polynomial having a remainder 6 when divided by (x-2)
Let's check:
Let x - 2 = 0 => x = 2
Putting in the above polynomial
[tex]p(x) = 2(2)^2+(2)-4\\Given \ that \ Remainder = 6\\6 = 2(4) +2-4\\6 = 8+2-4\\6 = 10-4\\6 = 6[/tex]
So, Proved that it has a remainder of 6 when divided by (x-2)
