Find a degree 3 polynomial p such that p(−1) = −3, p(0) =−2, p(1) = −1, and p(2) = 6.

Step-by-step explanation:

let polynomial P(x)=ax³+bx²+cx+d

-3=-a+b-c+d

-2=d

-1=a+b+c+d

-1=a+b+c-2

a+b+c=1

-3=-a+b-c-2

a-b+c=1

a+b+c-(a-b+c)=1-1=0

2b=0

b=0

so a+c=1

6=8a+4b+2c+d

6=8a+4b+2c-2

8a+4b+2c=8

4a+2b+c=4

or 4a+0+c=4

4a+c=4

a+c=1

subtract

3a=3

a=1

1+c=1

c=0

reqd .P(x)=1x³-2=x³-2

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.

c.

step-by-step explanation: