Question

Find the sum, difference or product.

$5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)$

Short answer

The solution$5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)=-3t^2+21t-22$

Step-by-step solution

Step 1. Apply the concept of addition and subtraction of polynomials.

We add and subtract polynomials using the properties of real numbers. The idea is to combine like terms (that is, terms with the same variables raised to the same powers) using the Distributive Property.

In subtracting polynomials, we have to remember that if a minus sign precedes an expression in parentheses, then the sign of every term within the parentheses is changed when we remove the parentheses.

Step 2. Combine like terms.

Given$5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)$

Let us combine the terms with same variables raised to the same powers.

$\begin{array}{l}5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)=15t-20-t^2-2-2t^2+6t\\ \Rightarrow 5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)=\left(-t^2-2t^2\right)+\left(15t+6t\right)-\left(20+2\right)\end{array}$

Step 3. Solve the expression.

From Step 2, we have:

$\begin{array}{l}5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)=\left(-t^2-2t^2\right)+\left(15t+6t\right)-\left(20+2\right)\\ \Rightarrow 5\left(3t-4\right)-\left(t^2+2\right)-2t\left(t-3\right)=-3t^2+21t-22\end{array}$