Question

How do you use synthetic substitution to factor `x^3+9x^2+24x+20` if x+5 is one of its factor?

Answer 1

`x^3+9x^2+24x+20 = color(blue)( (x+5)(x+2)(x+2))`

Explanation:

If `x+5` is a factor, your divisor in synthetic substitution is `-5`.

Step 1. Write only the coefficients of `x` in the dividend inside an upside-down division symbol.

`color(white)(Xll)|1color(white)(Xll)9color(white)(XX)24color(white)(XXl)20`
`color(white)(XX)|`
`color(white)(XX)stackrel("—————————————)`

Step 2. Put the divisor at the left.

`color(blue)(-5)|1color(white)(Xll)9color(white)(XX)24color(white)(XXl)20`
`color(white)(XX)|`
`color(white)(XX)stackrel("—————————————)`

Step 3. Drop the first coefficient of the dividend below the division symbol.

`-5|1color(white)(Xll)9color(white)(XX)24color(white)(XXl)20`
`color(white)(XX)|`
`color(white)(XX)stackrel("—————————————)`
`color(white)(Xll)|color(blue)(1)`

Step 4. Multiply the result by the divisor, and put the product in the next column.

`-5|1color(white)(Xll)9color(white)(XX)24color(white)(XXl)20`
`color(white)(Xll)|color(white)(Xl)color(blue)(-5)`
`color(white)(XX)stackrel("—————————————)`
`color(white)(Xll)|1`

Step 5. Add down the column.

`-5|1color(white)(Xll)9color(white)(XX)24color(white)(XXl)20`
`color(white)(Xll)|color(white)(ll)-5`
`color(white)(XX)stackrel("—————————————)`
`color(white)(Xll)|1" "color(white)(l)color(blue)(4)`

Step 6. Repeat Steps 4 and 5 until you can go no farther.

`-5|1color(white)(Xll)9color(white)(XX)24color(white)(XXl)20`
`color(white)(Xll)|color(white)(ll)-5color(white)(1)-20color(white)(1)-20`
`color(white)(XX)stackrel("————————————)`
`color(white)(Xll)|1" "color(white)(l)4" "color(white)(Xl)4" "color(white)(Xl)color(red)(0)`

∴ `(x^3+9x^2+24x+20)/(x+5) = x^2 + 4x +4`

You can factor the quadratic as `x^2 + 4x + 4 = (x+2)(x+2)`

So

`x^3+9x^2+24x+20 = (x+5)(x+2)(x+2)`