Answer:
\(\text{The remainder is 6 when }x^3-7x^2-18x+42\text{ is divided by (x + 3)}\)
Step-by-step explanation:
\(\text{Given the polynomial }x^3-7x^2-18x+42\)
we have to find the remainder when above polynomial is divided by (x+3)
\(P(x)=x^3-7x^2-18x+42\)
By remainder theorem,
\(P(3)=(-3)^3-7(-3)^2-18(-3)+42\)
\(P(3)=-27-63+54+42\)
\(P(x)=6\)
Hence, the remainder is 6
