How do you multiply (6a^2+ab-b^2)/(10a^2+5ab)*(2a^3+4a^2b)/(3a^2+5ab-2b^2)?

Jul 1, 2017

$\frac{6 {a}^{2} + a b - {b}^{2}}{10 {a}^{2} + 5 a b} \cdot \frac{2 {a}^{3} + 4 {a}^{2} b}{3 {a}^{2} + 5 a b - 2 {b}^{2}} \equiv \frac{2 a}{5}$

Explanation:

Let us denote the expression by $E$:

$E = \frac{6 {a}^{2} + a b - {b}^{2}}{10 {a}^{2} + 5 a b} \cdot \frac{2 {a}^{3} + 4 {a}^{2} b}{3 {a}^{2} + 5 a b - 2 {b}^{2}}$

We could multiply the expressions on the numerator and denominator but we would end up with high order polynomials which would be difficult to factorise and simplify

Portions of the expression can be immediately factorised as follows:

$E = \frac{6 {a}^{2} + a b - {b}^{2}}{5 a \left(2 a + b\right)} \cdot \frac{\left(2 {a}^{2}\right) \left(a + 2 b\right)}{3 {a}^{2} + 5 a b - 2 {b}^{2}}$

Then we cancel a factor of $a$ in the numerator and denominator:

$E = \frac{6 {a}^{2} + a b - {b}^{2}}{5 \left(2 a + b\right)} \cdot \frac{\left(2 a\right) \left(a + 2 b\right)}{3 {a}^{2} + 5 a b - 2 {b}^{2}}$

We can also factorise both of the quadratic expressions in the numerator and denominator:

$E = \frac{\left(3 a - b\right) \left(2 a + b\right)}{5 \left(2 a + b\right)} \cdot \frac{\left(2 a\right) \left(a + 2 b\right)}{\left(3 a - b\right) \left(a + 2 b\right)}$

Then we can can cancel the factors $\left(3 a - b\right)$, $\left(2 a + b\right)$, and $\left(a + 2 b\right)$ which all appear once in both the numerator and denominator, leaving:

$E = \frac{1}{5} \cdot \frac{2 a}{1}$

Thus we have:

$E = \frac{2 a}{5}$