Question

How do you divide `6x^3+5x^2-4x+4` by `2x+3`?

Answer 1

`3x^2-2x+1+1/(2x+3)`

Explanation:

`"one way is to use the divisor as a factor in the numerator"`

`"consider the numerator"`

`color(red)(3x^2)(2x+3)color(magenta)(-9x^2)+5x^2-4x+4`

`=color(red)(3x^2)(2x+3)color(red)(-2x)(2x+3)color(magenta)(+6x)-4x+4`

`=color(red)(3x^2)(2x+3)color(red)(-2x)(2x+3)color(red)(+1)(2x+3)color(magenta)(-3)+4`

`=color(red)(3x^2)(2x+3)color(red)(-2x)(2x+3)color(red)(+1)(2x+3)+1`

`"quotient "=color(red)(3x^2-2x+1)," remainder "=+1`

`rArr(6x^3+5x^2-4x+4)/(2x+3)=3x^2-2x+1+1/(2x+3)`

Answer 2

`3x^2-2x+1 + 1/(2x+3)`

Explanation:

For completeness sake, I'll demonstrate using synthetic division to do the same work. In some cases, the math can be easier to work through using synthetic division over traditional division, although if you are dividing by some term such as `ax+b` (where `a != 1`), there's an extra step you must remember at the end.

Start by taking the term you are dividing by and setting it equal to 0 to solve for the root in question:

`2x+3 = 0 => x = -3/2`

Write this number in a little boxed off area to the left of the line of paper, and then next to that boxed off area write down all of the coefficients of the polynomial, in descending exponent order. Ensure that any missing power of `x` is represented with a 0 coefficient in the line of coefficients you are writing. (This isn't the case in this problem, but worth noting anyway.)

`-3/2__|color(white)("aaaaa")6color(white)("aaaaa")5color(white)("aaaaa")-4color(white)("aaaaa")4`

Now, leave a little vertical space for a line of numbers you haven't written yet, and draw a horizontal line like you would for an addition problem:

`-3/2__|color(white)("aaaaa")6color(white)("aaaaa")5color(white)("aaaaa")-4color(white)("aaaaa")4`
`color(white)("aaaaaaaa")underline(color(white)("aaaaaaaaaaaaaaaaaaaaaaaaa"))`

Lastly, copy down the leading coefficient number below the line you drew. After this step you are set up and ready to begin:

`-3/2__|color(white)("aaaaa")6color(white)("aaaaa")5color(white)("aaaaa")-4color(white)("aaaaa")4`
`color(white)("aaaaaaaa")underline(color(white)("aaaaaaaaaaaaaaaaaaaaaaaaa"))`
`color(white)("aaaaaaaaaa")6`

Repeat the following steps over and over until you run out of numbers in the top row:

• Multiply the last number written underneath the line by the number in the "box" to the upper-left.
• Write this product just above the line underneath the next number to the right.
• Add the two numbers vertically in the next column to the right.
• Write this sum just under the line in the next column to the right.

Here's the work done. I've color coded each set as you go along:

`-3/2__|color(white)("aaaaa")6color(white)("aaaaa")5color(white)("aaaaa")-4color(white)("aaaaa")4`
`color(white)("aaaaaaaa")underline(color(white)("aaaaaa")color(green)(-9)color(white)("aaaaaaa")color(blue)(6)color(white)("aaaa")color(red)(-3)`

`color(white)("aaaaaaaaaa")6color(white)("aaa")color(green)(-4)color(white)("aaaaaaa")color(blue)(2)color(white)("aaaaa") color(red)(|__1)`

The last number boxed in red at the end of the bottom row of numbers is the remainder numerator, which should be placed over the original divided term of `2x+3`.

The other three numbers of the row represent, from right to left, the constant term, the `x` term, and the `x^2` term coefficients. However, as I noted at the beginning, when you use synthetic division to divide by a term of the form `ax+b` (`a != 1`), there's a final extra step: You must divide all of the coefficients at the end by the original divisor coefficient `a` (or, in this case, 2).

Thus, we divide the 6, -4, and 2 by 2 to get final coefficients of 3, -2, and 1. We can now write the final answer:

`3x^2-2x+1 + 1/(2x+3)`

Answer 3

`3x^2-2x+1+1/(2x+3)`

Explanation:

`color(white)("ddddddddddddddd")6x^3+5x^2-4x+4`
`color(magenta)(+3x^2)(2x+3)->color(white)("d")ul(6x^3+9x^2larr" Subtract" )`
`color(white)("ddddddddddddddd")0color(white)("d")-4x^2-4x+4`
`color(magenta)(-2x)(2x+3)->color(white)("ddddd")ul( -4x^2-6xlarr" Subtract")`
`color(white)("ddddddddddddddddddd.d")0color(white)("d")+2x+4`
`color(magenta)(+1)(2x+3)->color(white)("dddddddddddd..")ul(2x+3larr" Subtract")`
`color(white)("ddddddddddddddddddddddddddddd")color(magenta)( 1 larr" Remainder")`

Putting it all together:

`color(magenta)( 3x^2-2x+1+1/(2x+3) )`