Question

*For what ovalue is (x-5) a factor of x3+2x2 + cx + 10

Answer 1

Given that (x - 5) is a factor of

`f(x)=x^3+2x^2+cx+10`

Since x - 5 is a factor,

`\begin{gathered} x-5=0 \\ x=5 \end{gathered}`

x = 5 is one of the roots of the equation,

Substitute 5 for x into the eqaution to find c

`\begin{gathered} f(x)=x^3+2x^2+cx+10 \\ f(5)=5^3^{}+2(5)^2+c(5)+10=0 \\ 5^3+2(5)^2+c(5)+10=0 \\ 125+2(25)+5c+10=0 \\ 125+50+5c+10=0 \\ \text{Collect like terms} \\ 125+50+10+5c=0 \\ 185+5c=0 \\ 5c=-185 \\ \text{Divide both sides by 5} \\ (5c)/(5)=-(185)/(5) \\ c=-37 \end{gathered}`

Hence, the value of c for which (x - 5) is a factor is -37