Need some help factoring an expression
- Thread starter garfield
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How do I factor this expression? (x2 + 1) 1/2 + 2(x2 + 1)-1/2
I know that the answer is (x2 + 3) (x2 + 1) -1/2 but I don't know how to get to that answer. would appreciate help very much.
Thanks.
Just add them up:
\(\displaystyle (x^2+1)^{1/2}+2(x^2+1)^{-1/2}=\sqrt{x^2+1}+\dfrac{2}{\sqrt{x^2+1}}=\cdots\)
pka
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\(\displaystyle \left( {x^2 + 1} \right)^{1/2} + 2\left( {x^2 + 1} \right)^{ - 1/2}\)
\(\displaystyle \left( {x^2 + 1} \right)^{-1/2} \left[ {\left( {x^2 + 1} \right) + 2} \right]\)
\(\displaystyle \left( {x^2 + 1} \right)^{-1/2} \left[x^2+3 \right]\)
Hello, garfield!
pka is absolutely correct.
I'll try to explain the reasoning behind his factoring.
Suppose you are given \(\displaystyle a^5 + 2a^4\) to factor.
You'd factor out \(\displaystyle a^4\), wouldn't you? . \(\displaystyle a^4(a+2)\)
But let's think about how you came up with that game plan.
We have: .\(\displaystyle a^5 + 2a^4\)
We see that the two terms have some \(\displaystyle a\)'s in common.
. . How many?
Easy . . . It's the lesser of the two exponents, right?
. . So we'll factor out \(\displaystyle a^4\) from both terms.
Recall what it means to "factor out".
. . We are dividing out \(\displaystyle a^4\) from both terms.
If we wrote out all the baby-steps,
. . it would look this this: .\(\displaystyle a^4\left(a^{5-4} + 2a^{4-4}\right) \;=\;a^4(a + 2)\)
Back to the original problem: .\(\displaystyle (x^2+1)^{\frac{1}{2}} + 2(x^2+1)^{\text{-}\frac{1}{2}}\)
The two terms have some \(\displaystyle (x^2+1)\) in common.
. . How many?
Same rule: the lesser of the exponents.
. . So we will factor out \(\displaystyle (x^2+1)^{\text{-}\frac{1}{2}}\) from both terms.
Since "factor out" means divide out, we have:
. . \(\displaystyle (x^2+1)^{\text{-}\frac{1}{2}}\,\left[(x^2+1)^{\frac{1}{2}-(\text{-}\frac{1}{2})}} + 2(x^2+1)^{\text{-}\frac{1}{2}-(\text{-}\frac{1}{2})}\right] \)
\(\displaystyle =\;(x^2+1)^{-\frac{1}{2}}\,\left[(x^2+1)^1 + 2(x^1+1)^0\right]\)
\(\displaystyle =\;(x^2+1)^{-\frac{1}{2}}\,(x^2+1 + 2)\)
\(\displaystyle =\;(x^2+1)^{-\frac{1}{2}}(x^2+3)\)
pka is absolutely correct.
I'll try to explain the reasoning behind his factoring.
Suppose you are given \(\displaystyle a^5 + 2a^4\) to factor.
You'd factor out \(\displaystyle a^4\), wouldn't you? . \(\displaystyle a^4(a+2)\)
But let's think about how you came up with that game plan.
We have: .\(\displaystyle a^5 + 2a^4\)
We see that the two terms have some \(\displaystyle a\)'s in common.
. . How many?
Easy . . . It's the lesser of the two exponents, right?
. . So we'll factor out \(\displaystyle a^4\) from both terms.
Recall what it means to "factor out".
. . We are dividing out \(\displaystyle a^4\) from both terms.
If we wrote out all the baby-steps,
. . it would look this this: .\(\displaystyle a^4\left(a^{5-4} + 2a^{4-4}\right) \;=\;a^4(a + 2)\)
Back to the original problem: .\(\displaystyle (x^2+1)^{\frac{1}{2}} + 2(x^2+1)^{\text{-}\frac{1}{2}}\)
The two terms have some \(\displaystyle (x^2+1)\) in common.
. . How many?
Same rule: the lesser of the exponents.
. . So we will factor out \(\displaystyle (x^2+1)^{\text{-}\frac{1}{2}}\) from both terms.
Since "factor out" means divide out, we have:
. . \(\displaystyle (x^2+1)^{\text{-}\frac{1}{2}}\,\left[(x^2+1)^{\frac{1}{2}-(\text{-}\frac{1}{2})}} + 2(x^2+1)^{\text{-}\frac{1}{2}-(\text{-}\frac{1}{2})}\right] \)
\(\displaystyle =\;(x^2+1)^{-\frac{1}{2}}\,\left[(x^2+1)^1 + 2(x^1+1)^0\right]\)
\(\displaystyle =\;(x^2+1)^{-\frac{1}{2}}\,(x^2+1 + 2)\)
\(\displaystyle =\;(x^2+1)^{-\frac{1}{2}}(x^2+3)\)