How to solve 60 imes2555

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 60\times2555 | Equatix

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Question

$$60 \times 2555$$

Evaluate

$153300$

Steps For Long Multiplication

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

$$\begin{array}{c}\phantom{\times9999}60\\\underline{\times\phantom{99}2555}\\\end{array}$$

Now multiply the first number with the $1^{st}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 60 with 5. Write the result 300 at the end leaving 0 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}60\\\underline{\times\phantom{99}2555}\\\phantom{\times999}300\\\end{array}$$

Now multiply the first number with the $2^{nd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 60 with 5. Write the result 300 at the end leaving 1 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}60\\\underline{\times\phantom{99}2555}\\\phantom{\times999}300\\\phantom{\times99}300\phantom{9}\\\end{array}$$

Now multiply the first number with the $3^{rd}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 60 with 5. Write the result 300 at the end leaving 2 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}60\\\underline{\times\phantom{99}2555}\\\phantom{\times999}300\\\phantom{\times99}300\phantom{9}\\\phantom{\times9}300\phantom{99}\\\end{array}$$

Now multiply the first number with the $4^{th}$ digit in $2^{nd}$ number to get intermediate results. That is Multiply 60 with 2. Write the result 120 at the end leaving 3 spaces to the right like this.

$$\begin{array}{c}\phantom{\times9999}60\\\underline{\times\phantom{99}2555}\\\phantom{\times999}300\\\phantom{\times99}300\phantom{9}\\\phantom{\times9}300\phantom{99}\\\underline{\phantom{\times}120\phantom{999}}\\\end{array}$$

Now add the intermediate results to get final answer.

Show Solution

Factor

$2^{2}\times 3\times 5^{2}\times 7\times 73$