Question

Find all roots of f(x): f(x) = 6x^3-x^2-9x+4 if (2x-1) is a factor. Show all work.

Which procedure would you take first? Synthetic division?

And if its synthetic division, can you show how you divide by (2x-1)?

Answer 1

`x=(1)/(2),\;\;x=1,\;\;x=-(4)/(3)`

Explanation:

Given polynomial function:

`f(x) = 6x^3-x^2-9x+4`

If (2x - 1) is a factor of the given function, then 2x - 1 = 0.

Therefore:

`\implies 2x-1=0`

`\implies x=(1)/(2)`

To divide the given function by the factor (2x - 1), we can perform synthetic division.

As the leading coefficient of the divisor is not 1, divide the dividend and divisor by the leading coefficient of the divisor (2):

`\textsf{Dividend}: \quad 3x^3-(1)/(2)x^2-(9)/(2)x+2`

`\textsf{Divisor}: \quad \left(x-(1)/(2)\right)`

Perform Synthetic Division

Place ¹/₂ in the division box.

Write the coefficients of the dividend in descending order.

(Note: As no terms are missing, we do not need to use any zeros to fill in missing terms).

`\begin{array}c\vphantom{\frac12}(1)/(2) & 3 & -(1)/(2)& -(9)/(2) & 2\\\cline{1-1}\end{array}`

Bring the leading coefficient straight down:

`\begin{array}crrr\vphantom{\frac12}(1)/(2) & 3 & -(1)/(2)& -(9)/(2) & 2\\\cline{1-1}&\downarrow&&&\\\cline{2-5}&3\end{array}`

Multiply the number you brought down with the number in the division box and put the result in the next column (under the -¹/₂ ):

`\begin{array}crrr\vphantom{\frac12}(1)/(2) & 3 & -(1)/(2)& -(9)/(2) & 2\\\cline{1-1}\vphantom{\frac12}&\downarrow&(3)/(2)&&\\\cline{2-5}&3\end{array}`

Add the two numbers together and put the result in the bottom row:

`\begin{array}c\vphantom{\frac12}(1)/(2) & 3 & -(1)/(2)& -(9)/(2) & 2\\\cline{1-1}\vphantom{\frac12}&\downarrow&-(3)/(2)&&\\\cline{2-5}&3&1\end{array}`

Repeat:

`\begin{array}crrr\vphantom{\frac12}(1)/(2) & 3 & -(1)/(2)& -(9)/(2) & 2\\\cline{1-1}\vphantom{\frac12}&\downarrow&-(3)/(2)&(1)/(2)&\\\cline{2-5}&3&1&-4\end{array}`

`\begin{array}c\vphantom{\frac12}(1)/(2) & 3 & -(1)/(2)& -(9)/(2) & 2\\\cline{1-1}\vphantom{\frac12}&\downarrow&-(3)/(2)&(1)/(2)&-2\\\cline{2-5}&3&1&-4&0\end{array}`

The bottom row (except the last number) gives the coefficients of the quotient. The degree of the quotient is one less than that of the dividend.

The last number in the bottom row is the remainder.

Therefore, the factored function after synthetic division is:

`f(x)=(2x-1)(3x^2+x-4)`

The quadratic factor can be factored further:

`\implies 3x^2+4x-3x-4`

`\implies x(3x+4)-1(3x+4)`

`\implies (x-1)(3x+4)`

Therefore, the fully factored function is:

`f(x)=(2x-1)(x-1)(3x+4)`

To find the roots of the function, set each factor to zero and solve for x:

`2x-1=0 \implies x=(1)/(2)`

`x-1=0 \implies x=1`

`3x+4=0 \implies x=-(4)/(3)`

Therefore, the roots of the given function are:

`x=(1)/(2),\;\;x=1,\;\;x=-(4)/(3)`