Question

What is the following product? assume x > 0 4x sqrt 5x^2 + 2x^2 sqrt 6) ^2

Answer 1

Answer:

The product of the given expression is

 $(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=104x^4+16\sqrt{30}x^4$

or  

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=(104+16\sqrt{30})x^4$

Step-by-step explanation:

Given  that the expression is

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2$

Assume that x>0

To find the product of the given expression:

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2$

We may write it as

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=(4x\sqrt{5x^2}+2x^2\sqrt{6})\times (4x\sqrt{5x^2}+2x^2\sqrt{6})$

Now using the distributive property (each term in the expression is multiplied to each term in the another expression

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=(4x\sqrt{5}x+2x^2\sqrt{6})\times (4x\sqrt{5}x+2x^2\sqrt{6})$

$=(4x^2\sqrt{5}+2x^2\sqrt{6})\times (4x^2\sqrt{5}+2x^2\sqrt{6})$

$=(4x^2\sqrt{5})(4x^2\sqrt{5})+(4x^2\sqrt{5})(2x^2\sqrt{6})+(2x^2\sqrt{6})(4x^2\sqrt{5})+(2x^2\sqrt{6})(2x^2\sqrt{6})$

$=(4x^2\sqrt{5})^2+8x^4\sqrt{5}\sqrt{6}+8x^4\sqrt{6}\sqrt{5}+4x^4(6)$

$=16x^4(5)+16x^4\sqrt{5}\sqrt{6}+24x^4$

$=80x^4+16x^4\sqrt{5\times6}+24x^4$

$=80x^4+16x^4\sqrt{30}+24x^4$

$=80x^4+16\sqrt{30}x^4+24x^4$

Adding the like terms

$=104x^4+16\sqrt{30}x^4$

Therefore  

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=104x^4+16\sqrt{30}x^4$

Therefore the product of the given expression is

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=104x^4+16\sqrt{30}x^4$

or  

$(4x\sqrt{5x^2}+2x^2\sqrt{6})^2=(104+16\sqrt{30})x^4$