I am working on factoring \$a^4+b^4\$ and I have found two different solutions khổng lồ this.

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First, I have factored it khổng lồ \$\$a^4+b^4=(a+b)-2ab^3\$\$

But then I also found that the equation is factorable to lớn \$\$a^4+b^4=(a^2+b^2-sqrt2ab)(a^2+b^2+sqrt2ab)\$\$

First of all, why are there two ways of factoring this equation, và why are the solutions not the same?

Additionally, is it possible khổng lồ convert either solution so it is the same as the other? \$egingroup\$ Yeah I don't get why everyone is saying either of those is wrong. It's wrong to call the first one a factorization but the equation itself isn't wrong. The equation is true. We need to be more explicit about what we're calling "wrong" here. \$endgroup\$
Contrary lớn what"s being said in the comments, neither of those expressions is wrong because both yield \$a^4 + b^4\$ when expanded. It is, however, incorrect to gọi this a factorization:

\$\$a^4+b^4=(a+b)-2ab^3\$\$

When you factor an expression, say \$F(x)\$, you break it down into two or more factors, such as \$F(x) = G(x)H(x)K(x)\$. That is not what you did here. You broke it down into two factors plus a remainder of \$-2ab^3\$. That is not factoring.

To address the question more generally, since \$Bbb R\$ (or \$Bbb C\$, or \$Bbb Q\$, whichever you"re working with) is a field, then \$Bbb R\$ is a unique factorization tên miền (among other things, but those other things are irrelevant here). This means that every expression in \$Bbb R\$ has one và only one factorization (up khổng lồ ordering of the factors, for example \$x(x-1)\$ và \$(x-1)x\$ are considered the same factorization of \$x^2-x\$). An expression may have two factorizations that look different but actually aren"t.

For a simple example over \$Bbb C\$, we could say \$x^4 - 1 = (x^2 + 1)(x^2 - 1)\$ and we could say \$x^4 - 1 = (x^2 - x + i(1-x))(x^2 + x + i(1+x))\$. These look very different but they"re really the same. How can we be sure they"re the same? Break it into linear factors over \$Bbb C\$:\$\$ x^4 - 1 = (x-1)(x+1)(x+i)(x-i)\$\$

\$(x-1)(x+1) = x^2-1\$ and \$(x+i)(x-i) = x^2+1\$. That"s how we can get the first factorization.

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Alternatively, \$(x-1)(x-i) = x^2 - x + i(1-x)\$ và \$(x+1)(x+i) = x^2 + x + i(1+x)\$. That"s how we can get the second factorization.