Question

Divide f(x) by g(x) using long division and write the result using the division algorithm. Please help!

Answer 1

To use long division in polynomials, we follow steps similar to normal long division, but we look for the terms from higher degree to lower.

We are dividing:

$4x^3-7x^2+3$

By:

$2x-1$

We want to multiply the divisor by some expression that will make the higher term equal to the higher term of the dividend. The higher term of the dividend is 4x³, to get to that, we can multiply the divisor by 2x²:

$2x^2(2x-1)=2x^2\cdot2x-2x^2=4x^3-2x^2$

Now we have the same term, so we just substract what we have got from the dividend:

$\begin{gathered} 4x^3-7x^2+3 \\ -(4x^3-2x^2) \\ \\ 4x^3-4x^3-7x^2+2x^2+3 \\ -5x^2+3 \end{gathered}$

Notice that we don't have a third degree term anymore.

So, until now we have done:

- Multiplied the divisor by 2x²

- Got a remainder of -5x² + 3

Now, we just repeat with the remainder.

We want to multiply 2x - 1 so that the higher term is -5x², so we can multiply by -5x/2:

$-\frac{5x}{2}(2x-1)=-\frac{5x\cdot2x}{2}+\frac{5x}{2}=-5x^2+\frac{5x}{2}$

And we do the substraction:

$\begin{gathered} -5x^2+3 \\ -(-5x^2+\frac{5x}{2}) \\ \\ -5x^2+5x^2-\frac{5x}{2}+3 \\ -\frac{5x}{2}+3 \end{gathered}$

So, now we have got:

- Multiplied the divisor by 2x² and then by -5x/2

- Got a remainder of -5x/2 + 3

Now, we repeat once more:

To get -5x/2, we multiply the divisor by -5/4:

$-\frac{5}{4}(2x-1)=-\frac{5\cdot2x}{4}+\frac{5}{4}=-\frac{5x}{2}+\frac{5}{4}$

And we substract from the remainder:

$\begin{gathered} -\frac{5x}{2}+3 \\ -(-\frac{5x}{2}+\frac{5}{4}) \\ \\ -\frac{5x}{2}+\frac{5x}{2}+3-\frac{5}{4} \\ \frac{12-5}{4} \\ \frac{7}{4} \end{gathered}$

So, we have done:

- Multiplied the divisor by 2x² then -5x/2 then -5/4

- Got a remainder of 7/4

This means that th result of the division is:

$2x^2-\frac{5x}{2}-\frac{5}{4}$

And the remainder is:

$\frac{7}{4}$

But, the answer wants us to write what the dividend is equal to.

Let's write first in the division form:

$\frac{4x^3-7x^2+3}{2x-1}=2x^2-\frac{5x}{2}-\frac{5}{4}+\frac{\frac{7}{4}}{2x-1}$

Notice that we result is the quotient plus the remainder divided by the divisor.

If we multiply both sides by the divisor, we will get:

$4x^3-7x^2+3=(2x-1)\mleft(2x^2-\frac{5x}{2}-\frac{5}{4}\mright)+\frac{7}{4}$

That is the answer.