How to solve 256 imes5225

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 256\times5225 | Equatix

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Question

$$256 \times 5225$$

Evaluate

$1337600$

Steps For Long Multiplication

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

$$\begin{array}{c}\phantom{\times9999}256\\\underline{\times\phantom{999}5225}\\\end{array}$$

Now multiply the first number with the $1^{st}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 256 with 5. Write the result 1280 at the end leaving 0 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}256\\\underline{\times\phantom{999}5225}\\\phantom{\times999}1280\\\end{array}$$

Now multiply the first number with the $2^{nd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 256 with 2. Write the result 512 at the end leaving 1 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}256\\\underline{\times\phantom{999}5225}\\\phantom{\times999}1280\\\phantom{\times999}512\phantom{9}\\\end{array}$$

Now multiply the first number with the $3^{rd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 256 with 2. Write the result 512 at the end leaving 2 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}256\\\underline{\times\phantom{999}5225}\\\phantom{\times999}1280\\\phantom{\times999}512\phantom{9}\\\phantom{\times99}512\phantom{99}\\\end{array}$$

Now multiply the first number with the $4^{th}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 256 with 5. Write the result 1280 at the end leaving 3 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}256\\\underline{\times\phantom{999}5225}\\\phantom{\times999}1280\\\phantom{\times999}512\phantom{9}\\\phantom{\times99}512\phantom{99}\\\underline{\phantom{\times}1280\phantom{999}}\\\end{array}$$

Now add the intermediate results to get final answer.

Show Solution

Factor

$2^{8}\times 5^{2}\times 11\times 19$