Divide:
$\frac{x^2+6x+13}{x+3}$
The quotient is □
The remainder is □
We are tasked with dividing the quadratic polynomial \(P(x) = x^2 + 6x + 13\) by the linear binomial \(D(x) = x + 3\). According to the division algorithm for polynomials, there exist a unique quotient \(Q(x)\) and a remainder \(R\) such that \(P(x) = D(x)Q(x) + R\), where the degree of \(R\) is less than the degree of \(D(x)\). Since our divisor is of degree 1, the remainder must be a constant. We can perform this division using polynomial long division or algebraic manipulation.
To perform the division \(\frac{x^2 + 6x + 13}{x+3}\), we want to rewrite the numerator \(x^2 + 6x + 13\) in terms of the divisor \(x+3\). We observe that \(x^2 + 6x + 13\) can be grouped as follows:
\[x^2 + 6x + 13 = (x^2 + 3x) + (3x + 13).\]
We can factor out \(x\) from the first group:
\[x(x + 3) + (3x + 13).\]
Now, to isolate a factor of \((x+3)\) from the second part, we express \(3x + 13\) as \(3(x + 3) + 4\):
\[x(x + 3) + 3(x + 3) + 4.\]
Combining the terms with \((x+3)\):
\[(x+3)(x+3) + 4.\]
Thus, the expression becomes \(\frac{(x+3)(x+3) + 4}{x+3} = (x+3) + \frac{4}{x+3}\).
From our result in the previous step, the expression \(\frac{x^2 + 6x + 13}{x+3} = x+3 + \frac{4}{x+3}\) identifies the components of the division. The part that is not a fraction, \(x+3\), is the quotient \(Q(x)\). The constant left over in the numerator is the remainder \(R\). Therefore, the quotient is \(x+3\) and the remainder is \(4\).
Polynomial division is essential in calculus and algebra for simplifying complex expressions and identifying horizontal or slant asymptotes of rational functions. For example, when analyzing the behavior of the rational function \(f(x) = \frac{x^2 + 6x + 13}{x+3}\) as \(x\) approaches infinity, we see that \(f(x) \approx x+3\), indicating that the graph approaches the line \(y = x+3\) as \(x\) grows large. This technique helps engineers and mathematicians understand the long-term trends in systems modeled by rational functions.