phythagorean theorem?

Question

does (r+4)^2 simplify to r^2+8r+16

where does the 8r come from

Answer 1

(r+4)^2 is like saying (r+4) x (r+4). So using FOIL (firsts, outside,inner, and lasts) you get r x r=r^2; r x 4=4r; 4 x r=4r; 4 x 4=16. Then add the like terms (terms with same exponents and variables) to get r^2+8r+16. The 8r comes from the two 4r's that combined since they are like terms. But this isn't the phythagorean theorem, it is a^2 + b^2=c^2 and used in finding a length of a side in a right triangle.

Answer 2

You are actually looking at the FOIL method for factoring equations. FOIL stands for FIRST OUTER INNER LAST.

Break the equation down to (r+4)(r+4) and now multiply it. It gives you rxr or r^2 + 4r + 4r + 16. Combine like terms to get r^2+8r+16. In actuality, that is simplified to (r+4)^2 rather than how you stated it in your question which is the long way.

Answer 3

the 8r comes from multiplying 4r twice.
remember FOIL (first, outer, inner, last)
(r+4)(r+4) --> r^2 +4r +4r +16 --> r^2 + 8r +16

Answer 4

(r + 4)(r + 4)
= r(r + 4) + 4(r + 4)
= r^2 + 4r + 4r + 16
= r^2 + 8r + 16

Answer 5

Yes multiply (r + 4)*(r + 4)

So by foil you do the first which is r * r = r^2
Then inner which is r * 4 = 4r
Then outter which is r * 4 = 4r
Finally last which is 4 * 4 = 16
Add them all together and you get r^2 + 8r + 16

Answer 6

Yes. By using the FOIL method (First, Outter, Inner, Last).

(r + 4)*(r + 4) =

=r^2 (first) + 4r (outter) + 4r (inner) + 16 (last)

Combine like terms

=r^2 + 8r + 16

Answer 7

(r + 4)^2 means (r+4) (r+4)

you need to contribute
r*r + 4r + 4r + 4*4

r^2 + 8r + 16

that's how 8r came from.

Answer 8

=(r+4)(r+4)

r*r + 4r + 4r + 4*4

r^2 + 8r + 16

Answer 9

we initiate with a acceptable triangle with sides a, b, and c. We then build a great sq., out of four copies of our triangle. We finally end up with a sq., interior the midsection, with sides c (we are able to definitely prepare that that's a sq.). We now build a 2nd enormous sq., with comparable triangles that are arranged as interior the decrease part of the diagram. This sq. has the comparable section because of the fact the sq. above it. We now sum up the factors of the two enormous squares: section=2ab + c² section=2ab + a² + b² those 2 factors are equivalent: 2ab + c²=2ab + a² + b² c²=a² + b²