Question

divide. if there is a remainder, include it as a simplified fraction. \\((2z^{3} + 7z^{2} + 5z) \div (2z + 5)\\)

Explanation:

Step1: Factor the numerator

First, factor the polynomial in the numerator \(2z^{3}+7z^{2}+5z\). We can factor out a common factor of \(z\) first:
\(2z^{3}+7z^{2}+5z = z(2z^{2}+7z + 5)\)
Then, factor the quadratic \(2z^{2}+7z + 5\). We need two numbers that multiply to \(2\times5 = 10\) and add up to \(7\). Those numbers are \(2\) and \(5\). So we can rewrite the middle term:
\(2z^{2}+2z + 5z+5=2z(z + 1)+5(z + 1)=(2z + 5)(z + 1)\)
So the numerator becomes \(z(2z + 5)(z + 1)\)

Step2: Divide by the denominator

Now we have \(\frac{z(2z + 5)(z + 1)}{2z + 5}\). We can cancel out the common factor of \(2z + 5\) (assuming \(2z+5
eq0\)):
\(\frac{z(2z + 5)(z + 1)}{2z + 5}=z(z + 1)\)

Step3: Expand the result

Expand \(z(z + 1)\) using the distributive property:
\(z(z + 1)=z^{2}+z\)

Answer:

\(z^{2}+z\)