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factoring

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1-2p+p2+2rp-2rp2+r2p2 factors to (1-p-rp)2, but how? What is the process to obtaining that simplified form?

Feb 24, 2022

#1
+2

Choose one of the two variables and collect together like powers removing common factors on the way. So, choosing p,

\(\displaystyle 1-2p+p^{2}+2rp-2rp^{2}+r^{2}p^{2}\\=p^{2}(1-2r+r^{2})-2p(1-r)+1 \\ =p^{2}(1-r)^{2}-2p(1-r)+1.\)

If you don't see how to simplyfy that directly, let  p(1 - r) = X say, and you then have

\(\displaystyle X^{2}-2X+1 = (X-1)^{2}, \\ = \{p(1-r)-1\}^{2}.\)

Alternatively had you chosen r, you get

\(\displaystyle 1-2p+p^{2}+2rp-2rp^{2}+r^{2}p^{2} \\ =r^{2}p^{2}+2rp(1-p)+1-2p+p^{2}\\ = r^{2}p^{2}+2rp(1-p)+(1-p)^{2}.\)

This time, if needed, let rp = X and 1 - p = Y and you get

\(\displaystyle X^{2}+2XY+Y^{2}=(X+Y)^{2} \\=(rp+1-p)^{2}.\)

The two results differ by a negative sign within the brackets, but algebraically they're equivalent.

And btw, they are different from your given result, (which was incorrect).

Feb 24, 2022

#1
+2

Choose one of the two variables and collect together like powers removing common factors on the way. So, choosing p,

\(\displaystyle 1-2p+p^{2}+2rp-2rp^{2}+r^{2}p^{2}\\=p^{2}(1-2r+r^{2})-2p(1-r)+1 \\ =p^{2}(1-r)^{2}-2p(1-r)+1.\)

If you don't see how to simplyfy that directly, let  p(1 - r) = X say, and you then have

\(\displaystyle X^{2}-2X+1 = (X-1)^{2}, \\ = \{p(1-r)-1\}^{2}.\)

Alternatively had you chosen r, you get

\(\displaystyle 1-2p+p^{2}+2rp-2rp^{2}+r^{2}p^{2} \\ =r^{2}p^{2}+2rp(1-p)+1-2p+p^{2}\\ = r^{2}p^{2}+2rp(1-p)+(1-p)^{2}.\)

This time, if needed, let rp = X and 1 - p = Y and you get

\(\displaystyle X^{2}+2XY+Y^{2}=(X+Y)^{2} \\=(rp+1-p)^{2}.\)

The two results differ by a negative sign within the brackets, but algebraically they're equivalent.

And btw, they are different from your given result, (which was incorrect).

Guest Feb 24, 2022
#2
+153
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i posted as a guest, but thanks!

ChowMein  Mar 1, 2022