Question

19 \frac{4x^{2} - 2x + 6}{2x - 3}

Explanation:

To solve the division of the polynomial \(\frac{4x^2 - 2x + 6}{2x - 3}\), we can use polynomial long division.

Step 1: Divide the leading terms

Divide the leading term of the numerator \(4x^2\) by the leading term of the denominator \(2x\):
\[
\frac{4x^2}{2x}=2x
\]

Step 2: Multiply the divisor by the result

Multiply the divisor \(2x - 3\) by \(2x\):
\[
(2x - 3)\times2x = 4x^2 - 6x
\]

Step 3: Subtract this from the numerator

Subtract \(4x^2 - 6x\) from \(4x^2 - 2x + 6\):
\[
(4x^2 - 2x + 6)-(4x^2 - 6x)=4x^2 - 2x + 6 - 4x^2 + 6x = 4x + 6
\]

Step 4: Divide the new leading term

Now, divide the leading term of the new polynomial \(4x\) by the leading term of the divisor \(2x\):
\[
\frac{4x}{2x}=2
\]

Step 5: Multiply the divisor by this result

Multiply the divisor \(2x - 3\) by \(2\):
\[
(2x - 3)\times2 = 4x - 6
\]

Step 6: Subtract this from the previous remainder

Subtract \(4x - 6\) from \(4x + 6\):
\[
(4x + 6)-(4x - 6)=4x + 6 - 4x + 6 = 12
\]

So, when we divide \(4x^2 - 2x + 6\) by \(2x - 3\), we get a quotient of \(2x + 2\) and a remainder of \(12\). We can express this as:
\[
\frac{4x^2 - 2x + 6}{2x - 3}=2x + 2+\frac{12}{2x - 3}
\]

If we want to check if we can factor or simplify further, we note that the remainder is \(12\) and the divisor is \(2x - 3\), so the division is complete in terms of polynomial long division.

If we were to find the value of \(x\) that makes the division exact (i.e., remainder \(0\)), we set the remainder equal to \(0\):
\[
12 = 0
\]
which is not possible, so there is no real value of \(x\) that makes the division exact. However, if we were just performing the polynomial division, the result is \(2x + 2+\frac{12}{2x - 3}\).

Answer:

To solve the division of the polynomial \(\frac{4x^2 - 2x + 6}{2x - 3}\), we can use polynomial long division.

Step 1: Divide the leading terms

Divide the leading term of the numerator \(4x^2\) by the leading term of the denominator \(2x\):
\[
\frac{4x^2}{2x}=2x
\]

Step 2: Multiply the divisor by the result

Multiply the divisor \(2x - 3\) by \(2x\):
\[
(2x - 3)\times2x = 4x^2 - 6x
\]

Step 3: Subtract this from the numerator

Subtract \(4x^2 - 6x\) from \(4x^2 - 2x + 6\):
\[
(4x^2 - 2x + 6)-(4x^2 - 6x)=4x^2 - 2x + 6 - 4x^2 + 6x = 4x + 6
\]

Step 4: Divide the new leading term

Now, divide the leading term of the new polynomial \(4x\) by the leading term of the divisor \(2x\):
\[
\frac{4x}{2x}=2
\]

Step 5: Multiply the divisor by this result

Multiply the divisor \(2x - 3\) by \(2\):
\[
(2x - 3)\times2 = 4x - 6
\]

Step 6: Subtract this from the previous remainder

Subtract \(4x - 6\) from \(4x + 6\):
\[
(4x + 6)-(4x - 6)=4x + 6 - 4x + 6 = 12
\]

So, when we divide \(4x^2 - 2x + 6\) by \(2x - 3\), we get a quotient of \(2x + 2\) and a remainder of \(12\). We can express this as:
\[
\frac{4x^2 - 2x + 6}{2x - 3}=2x + 2+\frac{12}{2x - 3}
\]

If we want to check if we can factor or simplify further, we note that the remainder is \(12\) and the divisor is \(2x - 3\), so the division is complete in terms of polynomial long division.

If we were to find the value of \(x\) that makes the division exact (i.e., remainder \(0\)), we set the remainder equal to \(0\):
\[
12 = 0
\]
which is not possible, so there is no real value of \(x\) that makes the division exact. However, if we were just performing the polynomial division, the result is \(2x + 2+\frac{12}{2x - 3}\).