Question

3-sqrt(x)/9-x. how do you break down?

Answer 1

\[\frac{3-\sqrt x}{9-x}\]

factorize the denominator
use \[(a+b)(a-b)=a^2-ab+ba-b^2=a^2-b^2\]

cancel common factors

Answer 2

yes i know that but how do you that for sqrt?

Answer 3

use \(x=\sqrt x^2\)

Answer 4

mmm so 3-sqet(x)=(x^2-3^9)?

Answer 5

your looking at the numerator (bottom bit of the fraction)

you should be looking at the denominator ( the bottom bit of the fraction)
\[9-x=(\cdots)(\cdots)\]

Answer 6

9-x...... sqert(3)-(x)??

Answer 7

\[9-x=3^2-\sqrt x^2=(\cdots)(\cdots)\]

Answer 8

(3-sq\[\left( 3-\sqrt{x} \right)\]

Answer 9

(3-sq\[\left( 3-\sqrt{x} \right)\](3-sq\[\left( 3-\sqrt{x} \right)\]?

Answer 10

\[\left( 3-\sqrt{x} \right)\]

Answer 11

\left( 3-sqrt{x} \right) \left( 3-sqrt{x} \right)

Answer 12

(3-sqrt(x)) (3-sqrt(x)

Answer 13

not quite\[a^2-b^2=(a+b)(a-b)\]

so\[9-x=3^2-\sqrt x^2=\]

Answer 14

3-sqrt(x)?

Answer 15

hope its right

Answer 16

\[9-x=3^2-\sqrt x^2=(3+\sqrt x)(3-\sqrt x)\]

Answer 17

so\[\frac{3-\sqrt x}{9-x}=\frac{3-\sqrt x}{(3+\sqrt x)(3-{\sqrt x)}}\]\[\qquad\qquad =\frac{\cancel{3-\sqrt x}}{(3+\sqrt x)\cancel{(3-{\sqrt x)}}}\]

Answer 18

Thanks!!

Answer 19

what about x^3+64/x+4

Answer 20

I got　(x+4)(x^2+4x+16)/(x+4)?

Answer 21

almost

\[(a^3+b^3)=(a+b)(a^2-ab+b^2)\]

Answer 22

once you have fixed the sign error, cancel the common factor

Answer 23

Thank you it was very helpful!!