Question

What is the following simplified product? Assume x greater-than-or-equal-to 0 (StartRoot 10 x Superscript 4 Baseline EndRoot minus x StarRoot 5 x squared EndRoot) (2 StartRoot 15 x Superscript 4 Baseline EndRoot + StartRoot 3 x cubed EndRoot) 10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot + x squared StartRoot 15 x EndRoot 11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus x Superscript 4 Baseline StartRoot 75 EndRoot + x squared StartRoot 15 EndRoot 10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x squared StartRoot 15 EndRoot 11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x cubed StartRoot 15 x EndRoot

Answer 1

Answer:

$\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$

Step-by-step explanation:

To find:

Simplified product of:

$(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})$

Solution:

First of all, let us have a look at some of the formula:

1. $(a+b) (c+d) = ac+bc+ad+bd$

2. $a^b\times a^c =a^{b+c }$

3. $\sqrt{a^{2b}} = \sqrt{a^b.a^b}=a^b$

4. $\sqrt a \times \sqrt b = \sqrt{a\times b}$

Now, let us apply the above formula to solve the given expression.

$(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\\Rightarrow(\sqrt{10x^4})(2\sqrt{15x^4})+(\sqrt{10x^4})(\sqrt{3x^3})-(x\sqrt{5x^2})(2\sqrt{15x^4})-(x\sqrt{5x^2})(\sqrt{3x^3})\\\\\Rightarrow2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\\Rightarrow\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$

The answer is:

$\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$