QUESTION IMAGE
Question
question
examples
use the long division method to find the
result when
$6x^4 - 22x^3 - 13x^2 - 23x - 11$ is
divided by $2x^2 - 8x - 3$. if there is a
remainder, express the result in the
form $q(x) + \frac{r(x)}{b(x)}$.
Step1: Divide the leading terms
Divide the leading term of the dividend \(6x^4\) by the leading term of the divisor \(2x^2\). So, \(\frac{6x^4}{2x^2}=3x^2\). This is the first term of the quotient \(q(x)\).
Step2: Multiply and subtract
Multiply the divisor \(2x^2 - 8x - 3\) by \(3x^2\): \(3x^2(2x^2 - 8x - 3)=6x^4-24x^3 - 9x^2\).
Subtract this from the dividend: \((6x^4 - 22x^3 - 13x^2 - 23x - 11)-(6x^4 - 24x^3 - 9x^2)= (6x^4-6x^4)+(-22x^3 + 24x^3)+(-13x^2 + 9x^2)-23x - 11 = 2x^3 - 4x^2 - 23x - 11\).
Step3: Divide the new leading terms
Now, divide the leading term of the new dividend \(2x^3\) by the leading term of the divisor \(2x^2\). So, \(\frac{2x^3}{2x^2}=x\). This is the next term of the quotient.
Step4: Multiply and subtract again
Multiply the divisor \(2x^2 - 8x - 3\) by \(x\): \(x(2x^2 - 8x - 3)=2x^3 - 8x^2 - 3x\).
Subtract this from the new dividend \(2x^3 - 4x^2 - 23x - 11\): \((2x^3 - 4x^2 - 23x - 11)-(2x^3 - 8x^2 - 3x)=(2x^3-2x^3)+(-4x^2 + 8x^2)+(-23x + 3x)-11 = 4x^2 - 20x - 11\).
Step5: Divide the new leading terms again
Divide the leading term of the new dividend \(4x^2\) by the leading term of the divisor \(2x^2\). So, \(\frac{4x^2}{2x^2}=2\). This is the next term of the quotient.
Step6: Multiply and subtract the last time
Multiply the divisor \(2x^2 - 8x - 3\) by \(2\): \(2(2x^2 - 8x - 3)=4x^2 - 16x - 6\).
Subtract this from the new dividend \(4x^2 - 20x - 11\): \((4x^2 - 20x - 11)-(4x^2 - 16x - 6)=(4x^2-4x^2)+(-20x + 16x)+(-11 + 6)= - 4x - 5\).
The remainder \(r(x)=-4x - 5\) and the divisor \(b(x)=2x^2 - 8x - 3\), and the quotient \(q(x)=3x^2+x + 2\).
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
\(3x^2 + x + 2+\frac{-4x - 5}{2x^2 - 8x - 3}\) (or \(3x^2 + x + 2+\frac{-4x - 5}{2x^2 - 8x - 3}\) can also be written as \(3x^2 + x + 2-\frac{4x + 5}{2x^2 - 8x - 3}\) by factoring out the negative sign from the numerator)