# Step-by-step Solution

## Multiply $\left(t+1\right)\left(3t-1\right)^2$

Go!
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### Videos

$9t^2t+t^2\cdot 9-6tt-6t+t+1$

## Step-by-step explanation

Problem to solve:

$\left(t\:+\:1\right)\left(3t\:-\:1\right)^2$
1

A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$

• Square of the first term: $\left(3t\right)^2 = \left(3t\right)^2$
• Double product of the first by the second: $2\left(3t\right)\left(-1\right) = 2\cdot 3t\cdot -1$
• Square of the second term: $\left(-1\right)^2 = {\left(-1\right)}^2$

$9t^2t+t^2\cdot 9-6tt-6t+t+1$

$9t^2t+t^2\cdot 9-6tt-6t+t+1$

### Problem Analysis

$\left(t\:+\:1\right)\left(3t\:-\:1\right)^2$

### Main topic:

Multiplication of numbers

~ 0.05 seconds