What is two ways to rewrite 21x^2 -42x +14 in equvalent form? i really need help here lol-

7(3x^2-6x+2)

Step-by-step explanation:

21, 42, 14 they can be divided by 7 so we put 7 out

we know that

the volume of the oblique pyramid with a square base is equal to

V=\frac{1}{3}BhV=

3

1

Bh

where

B is the area of the base

h is the height of the pyramid

In this problem

The area of the base is

B=EF^{2}\ units^{2}B=EF

2

units

2

h=AB\ unitsh=AB units

The two formulas to calculate h are

Applying the Pythagoras Theorem

h^{2}=AC^{2}-CB^{2}h

2

=AC

2

−CB

2

> h=\sqrt{AC^{2}-CB^{2}}h=

AC

2

−CB

2

h^{2}=AD^{2}-DB^{2}h

2

=AD

2

−DB

2

> h=\sqrt{AD^{2}-DB^{2}}h=

AD

2

−DB

2

therefore

the answer is

AB and EF

AC, CB and EF

AD, DB and EF