2x^(3)+3x^(2)-21x+20 divided by x-2 if there is a remainder express the result is the form of q(x)+(r(x))/(b(x)) explain simply and without all the brackets pleae

1 answer

To divide the polynomial \(2x^3 + 3x^2 - 21x + 20\) by \(x - 2\), we can use polynomial long division.

  1. Setup the Division: We want to divide \(2x^3 + 3x^2 - 21x + 20\) by \(x - 2\).

  2. Divide the Leading Terms: Divide the leading term of the dividend \(2x^3\) by the leading term of the divisor \(x\): \[ 2x^2 \] This will be the first term of the quotient.

  3. Multiply and Subtract: Multiply \(2x^2\) by \(x - 2\): \[ 2x^2(x - 2) = 2x^3 - 4x^2 \] Now subtract this from the original polynomial: \[ (2x^3 + 3x^2) - (2x^3 - 4x^2) = 3x^2 + 4x^2 = 7x^2 \] Bring down the next term, which is \(-21x\): \[ 7x^2 - 21x \]

  4. Repeat the Process: Now divide \(7x^2\) by \(x\): \[ 7x \] Multiply \(7x\) by \(x - 2\): \[ 7x(x - 2) = 7x^2 - 14x \] Subtract: \[ (7x^2 - 21x) - (7x^2 - 14x) = -21x + 14x = -7x \] Bring down the next term (constant 20): \[ -7x + 20 \]

  5. Divide Again: Now divide \(-7x\) by \(x\): \[ -7 \] Multiply \(-7\) by \(x - 2\): \[ -7(x - 2) = -7x + 14 \] Subtract: \[ (-7x + 20) - (-7x + 14) = 20 - 14 = 6 \]

  6. Final Result: Now we have determined the quotient and the remainder:

    • Quotient \(q(x) = 2x^2 + 7x - 7\)
    • Remainder \(r(x) = 6\)

Putting it all together, we can express the result as: \[ 2x^2 + 7x - 7 + \frac{6}{x - 2} \]

So, in the requested form, it's: \[ q(x) + \frac{r(x)}{b(x)} = 2x^2 + 7x - 7 + \frac{6}{x - 2} \]