Question

divide. if the polynomial does not divide evenly, include the remainder as a fraction. \\((3d^4 - 110d^2 + 12d) \div (d - 6)\\)

Explanation:

Step1: Use Polynomial Long Division

We divide \(3d^4 - 110d^2 + 12d\) by \(d - 6\). First, divide the leading term \(3d^4\) by \(d\) to get \(3d^3\). Multiply \(d - 6\) by \(3d^3\): \(3d^3(d - 6)=3d^4 - 18d^3\). Subtract this from the original polynomial: \((3d^4 - 110d^2 + 12d)-(3d^4 - 18d^3)=18d^3 - 110d^2 + 12d\).

Step2: Next Term in Quotient

Divide \(18d^3\) by \(d\) to get \(18d^2\). Multiply \(d - 6\) by \(18d^2\): \(18d^2(d - 6)=18d^3 - 108d^2\). Subtract: \((18d^3 - 110d^2 + 12d)-(18d^3 - 108d^2)= - 2d^2 + 12d\).

Step3: Next Term in Quotient

Divide \(-2d^2\) by \(d\) to get \(-2d\). Multiply \(d - 6\) by \(-2d\): \(-2d(d - 6)= - 2d^2 + 12d\). Subtract: \((-2d^2 + 12d)-(-2d^2 + 12d)=0\). Wait, but let's check the original polynomial again. Wait, the original polynomial is \(3d^4 - 110d^2 + 12d\), maybe I made a mistake. Wait, no, let's re - express the dividend: \(3d^4+0d^3 - 110d^2 + 12d+0\).

Wait, let's do it properly. Let's use synthetic division or check if \(d = 6\) is a root (Remainder Theorem). Let \(f(d)=3d^4 - 110d^2 + 12d\). Then \(f(6)=3\times6^4-110\times6^2 + 12\times6=3\times1296-110\times36 + 72=3888-3960 + 72=0\). So \(d - 6\) is a factor.

Now, perform polynomial division:

\(3d^4 - 110d^2 + 12d=(d - 6)(3d^3 + 18d^2 - 2d)\)

Wait, let's multiply \((d - 6)(3d^3 + 18d^2 - 2d)=3d^4+18d^3-2d^2-18d^3 - 108d^2 + 12d=3d^4-110d^2 + 12d\). Yes, that works.

Answer:

\(3d^3 + 18d^2 - 2d\)