Question

1. ((x^4 + 3x^3 - 4x^2 + 5x + 3) div (x^2 + x + 4))

Explanation:

Looking at the problem, we have a polynomial division: \((x^{4}+3x^{3}-4x^{2}+5x + 3)\div(x^{2}+x + 4)\)

Step 1: Divide the leading terms

Divide the leading term of the dividend \(x^{4}\) by the leading term of the divisor \(x^{2}\):
\(\frac{x^{4}}{x^{2}}=x^{2}\)

Step 2: Multiply the divisor by \(x^{2}\)

Multiply \((x^{2}+x + 4)\) by \(x^{2}\):
\(x^{2}(x^{2}+x + 4)=x^{4}+x^{3}+4x^{2}\)

Step 3: Subtract this from the dividend

Subtract \((x^{4}+x^{3}+4x^{2})\) from \((x^{4}+3x^{3}-4x^{2}+5x + 3)\):

$$
LATEXBLOCK0
$$

Step 4: Divide the new leading term

Divide the leading term of \(2x^{3}-8x^{2}+5x + 3\) (which is \(2x^{3}\)) by the leading term of the divisor (\(x^{2}\)):
\(\frac{2x^{3}}{x^{2}} = 2x\)

Step 5: Multiply the divisor by \(2x\)

Multiply \((x^{2}+x + 4)\) by \(2x\):
\(2x(x^{2}+x + 4)=2x^{3}+2x^{2}+8x\)

Step 6: Subtract this from the new dividend

Subtract \((2x^{3}+2x^{2}+8x)\) from \(2x^{3}-8x^{2}+5x + 3\):

$$
LATEXBLOCK1
$$

Step 7: Divide the new leading term

Divide the leading term of \(-10x^{2}-3x + 3\) (which is \(-10x^{2}\)) by the leading term of the divisor (\(x^{2}\)):
\(\frac{-10x^{2}}{x^{2}}=- 10\)

Step 8: Multiply the divisor by \(-10\)

Multiply \((x^{2}+x + 4)\) by \(-10\):
\(-10(x^{2}+x + 4)=-10x^{2}-10x - 40\)

Step 9: Subtract this from the new dividend

Subtract \((-10x^{2}-10x - 40)\) from \(-10x^{2}-3x + 3\):

$$
LATEXBLOCK2
$$

Since the degree of the remainder \(7x + 43\) is less than the degree of the divisor \(x^{2}+x + 4\), we stop here.

The quotient is \(x^{2}+2x - 10\) and the remainder is \(7x + 43\)

So, \((x^{4}+3x^{3}-4x^{2}+5x + 3)\div(x^{2}+x + 4)=x^{2}+2x - 10+\frac{7x + 43}{x^{2}+x + 4}\)

Answer:

The quotient is \(x^{2}+2x - 10\) and the remainder is \(7x + 43\) (or in the form \(x^{2}+2x - 10+\frac{7x + 43}{x^{2}+x + 4}\))