Is the HCF of a^3-8, a^4+16 and a^3+2a^2+4a just 1? Please answer........

1 answer:

3 0

First we find the factors
a^3-8 is the differnce between 2 perfect cubes and is factored out to
(a-2)(a^2+2a+4)
the factors are (a-2) and (a^2+2a+4)

a^4+16
this is not factorable, unless you mistiped it and it was really a^4-16
the factor is 1

a^3+2a^2+4a
factor out the a
a(a^2+2a+4)
this is the most factored out form
the factors are (a) and (a^2+2a+4)

there are no common factors except 1 so 1 is the answer

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Answer:

B'(16,14)

Step-by-step explanation:

In square ABCD, A(2,7) C(8,1) D(2,1). First, find the coordinates of the vertex B. From the attached diagram you can see that CD=6 units and is parallel to the x-axis, AD=6 units and is parallel to the y-axis. So, B(8,7).

The origin O is the center of dilation, 2 is the scale factor. The rule of the dilation is

(x,y)→(2x,2y)

Thus,

B(8,7)→B'(16,14) (see attached figure).