Question

F(x)= 5x^3+37x^2+11x-21 find the zeros of f(x+5)

Answer 1

Answer:
`-12, -6, -(22)/(3)`

Explanation:

Note that f(x+5) represents the graph of f(x) shifted 5 units left. So, we can solve f(x)=0 and then subtract 5 from all of the solutions.

`5x^3 +37x^2 +11x-21=0`

We note by inspection that
`x=-1` is a root.

`\longrightarrow (5x^3 +37x^2 +11x-21)/(x+1)=(5x^2 (x+1)+32x(x+1)-21(x+1))/(x+1)=5x^2 +32x-21\\\\\implies (x+1)(5x^2 +32x-21)=0\\\\(x+1)(5x-3)(x+7)=0\\\\x=-7, -1, (3)/(5)`

Subtracting 5 from our solutions, we get
`-12, -6, -(22)/(3)`.