Question

How do you use synthetic substitution to evaluate f(2) for `f(x)=x^3+2x^2-4x-5`?

Answer 1

`f(2)=3`

Explanation:

`f(x)=x^3+2x^2-4x-5`
When the problem asks to evaluate f(2), simply substitute in 2 for every time you see "x" in the function. So we have:
`f(2)=(2)^3+2(2)^2-4(2)-5`
Now we simplify each term:
`f(2)=8+2(4)-8-5`
`f(2)=8+8-8-5`
`f(2)=3`

Answer 2

`f(2)=3`

Explanation:

According to Remainder theorem, if a polynomial `f(x)` is divided by a monomial of degree one say `x-a`, the remainder is `f(a)`.

Hence, to evaluate `f(2)` for `x^3+2x^2-4x-5`, we should divided it by `x-2`

One Write the coefficients of `x` in the dividend inside an upside-down division symbol.

`color(white)(1)|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5`
`color(white)(1)|" "color(white)(X)`
`" "stackrel("—————————————)`

Two Put the divisor at the left.

`2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5`
`color(white)(1)|" "color(white)(X)`
`" "stackrel("—————————————)`

Three Drop the first coefficient of the dividend below the division symbol.

`2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5`
`color(white)(1)|" "color(white)(X)`
`" "stackrel("—————————————)`
`color(white)(1)|color(white)(X)color(red)1`

Four Multiply the result by the constant, and put the product in the next column.

`2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5`
`color(white)(1)|" "color(white)(XX1)2`
`" "stackrel("—————————————)`
`color(white)(1)|color(white)(X)color(blue)1`

Five Add down the column.

`2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5`
`color(white)(1)|" "color(white)(XX1)2`
`" "stackrel("—————————————)`
`color(white)(1)|color(white)(X)color(blue)1color(white)(X11)color(red)4`

Six Repeat Steps Four and Five until you can go no farther.

`2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5`
`color(white)(1)|" "color(white)(XX1)2color(white)(XXXX)8color(white)(XXxx)8`
`" "stackrel("—————————————)`
`color(white)(1)|color(white)(X)color(blue)1color(white)(X11)color(red)4color(white)(XXXx)color(red)4color(white)(XXXX)color(red)3`

Remainder is `3`. Hence `f(2)=3`