Factoring given expression: $3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}$

Question

Factor given expression

$$3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}$$

It seems that there's $x^{\frac 1 2 }$ in common. Let's rewrite this expression

$$3x^{3 \frac{1}{2}}-9x^{\frac 1 2}+6x^{-\frac {1}{2} }$$

Recalling that $x^{\frac 1 2 } = t $

$$3t^3-9t+6t^{-1}$$

Where have I gone wrong so far?

Answer 1

$$3t^3 -9t +6t^{-1} =(3t^4 -9t^2 +6)t^{-1} .$$ For the parenthesis, let $u=t^2$; can you factor it?

Answer 2

Or $$\frac{3(t^4-3t^2+2)}{t}$$ or $$\frac{3(t-\sqrt2)(t+\sqrt2)(t-1)(t+1)}{t}.$$

Answer 3

let's solve it from where you left. Write the last expression as

$$\frac{3t^4-9t^2+6}{t}$$

now factorize the numerator which gives us

$$3*(t^2-1)*(t^2-2)$$

thus expression becomes

$$\frac{3*(t^2-1)*(t^2-2)}{t}$$

put $t=x^{1/2}$

Answer 4

$$3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}=3x^{-\frac{1}{2}}(x^2-3x+2)= 3x^{-\frac{1}{2}}(x-2)(x-1) $$ As you see, there is no need for substitution.

Answer 5

It seems that there's $x^{\frac12}$ in common.

No, actually the smallest term $x^{-\frac12}$ is a common term. Then: $$3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}=3x^{-\frac12}(x^2-3x+2)=3x^{-\frac12}(x-1)(x-2).$$ Now, if you want, you can use $x^{-\frac12}=t$ to get rid of the fractional exponent.