Question

Divide using polynomial long division.

(3x²+x-7)%(x+3)

Answer 1

Dividing
`3x^2+x-7 by x+3` using polynomial long division leaves a remainder of x-7. So,
`(3x^2+x-7)%(x+3) = x-7.`

Polynomial Long Division for
`(3x²+x-7)%(x+3)`

Step 1: Set up the division

x+3

`| 3x^2 + x - 7`

-------

`3x^2 + 9x`

-8x

-7

Step 2: Divide the leading term of the dividend by the leading term of the divisor

- Bring down the first term of the dividend
`(3x^2)` to the quotient.

x

`| 3x^2 + x - 7`

-------

`3x^2 + 9x`

Step 3: Multiply the divisor by the quotient and subtract from the dividend

- Multiply the divisor (x+3) by the quotient (x).

- Subtract the product
`(3x^2 + 9x)` from the dividend.

x

`| 3x^2 + x - 7`

-------

`3x^2 + 9x`

-8x - 7

Step 4: Bring down the next term of the dividend and update the divisor

- Bring down the next term of the dividend (x) to the remainder.

- Update the divisor by attaching the remainder (-8x) to it.

x

`| 3x^2 + x - 7`

-------

`3x^2 + 9x + (-8x)`

-7

x + 3

Step 5: Repeat steps 2-4 until the remainder is a constant term or a polynomial of lower degree than the divisor

- Since the divisor has a higher degree than the remaining polynomial (-7), the remainder is a constant term.

- Therefore, the division is complete.

Answer:

`(3x²+x-7)%(x+3) = x - 7`

Note: The symbol "%" represents the modulo operation, which gives the remainder when dividing two polynomials.