What is the complete factorization of 4x^2-9y^4?

1 answer:

7 0

Use the difference of squares factor rule to get
4x^2-9y^4
(2x)^2-(3y^2)^2
(2x-3y^2)(2x+3y^2)

The final answer is choice C
Note: the order of the factors does not matter

You might be interested in

What's 1+1 ??? enjoy!

Natalija [7]

1+1= -2628474
hope it helps

5 0

2 years ago

Find each absolute value | 4+2i | | 5-i | | -3i |

galina1969 [7]

How to get answer for number 1: | 4+2i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=4,\:b=2\\=\sqrt{4^2+2^2}\\Refine\\=\sqrt{20}\\\sqrt{20}=2\sqrt{5}\\=2\sqrt{5}$

How to get answer for number 2: | 5-i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=5,\:b=-1\\=\sqrt{5^2+\left(-1\right)^2}\\=\sqrt{26}$

Number 3 how to get answer: | -3i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=0,\:b=-3\\=\sqrt{0^2+\left(-3\right)^2}\\Refine\\=\sqrt{9}\\\sqrt{9}\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=\sqrt{3^2}\\\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a\\\sqrt{3^2}=3\\= 3$