How to solve 3812 imes258

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 3812\times258 | Equatix

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Question

$$3812 \times 258$$

Evaluate

$983496$

Steps For Long Multiplication

First line up the numbers vertically and match the places from the right like this.

$$\begin{array}{c}\phantom{\times99}3812\\\underline{\times\phantom{999}258}\\\end{array}$$

Now multiply the first number with the $1^{st}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 3812 with 8. Write the result 30496 at the end leaving 0 spaces to the right like this.

$$\begin{array}{c}\phantom{\times99}3812\\\underline{\times\phantom{999}258}\\\phantom{\times9}30496\\\end{array}$$

Now multiply the first number with the $2^{nd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 3812 with 5. Write the result 19060 at the end leaving 1 spaces to the right like this.

$$\begin{array}{c}\phantom{\times99}3812\\\underline{\times\phantom{999}258}\\\phantom{\times9}30496\\\phantom{\times}19060\phantom{9}\\\end{array}$$

Now multiply the first number with the $3^{rd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 3812 with 2. Write the result 7624 at the end leaving 2 spaces to the right like this.

$$\begin{array}{c}\phantom{\times99}3812\\\underline{\times\phantom{999}258}\\\phantom{\times9}30496\\\phantom{\times}19060\phantom{9}\\\underline{\phantom{\times}7624\phantom{99}}\\\end{array}$$

Now add the intermediate results to get final answer.

$$\begin{array}{c}\phantom{\times99}3812\\\underline{\times\phantom{999}258}\\\phantom{\times9}30496\\\phantom{\times}19060\phantom{9}\\\underline{\phantom{\times}7624\phantom{99}}\\\phantom{\times}983496\end{array}$$

Show Solution

Factor

$2^{3}\times 3\times 43\times 953$