Rationalizing the numerator and complex factoring

**skg94** · February 6th 2013, 06:23 PM

1. [ 1 / sqroot(x)] - 1 / (x-1)
2. [x*sqroot(x)] -8 / (x-4)

Factor:
a) x^6+8
b)x^4-16
c)r^8-1
d) x^3-x^2-16x+16
f)x^4-3x^3-7x^2+27x-18

are there any easy ways to factor theses without graphing, algebraically. I only know of synthetic division.

**ibdutt** · February 6th 2013, 07:34 PM

There are certain formulas which one must learn. (a^3+b^3) = (a+b)(a^2-ab+b^2); (a^3-b^3) = (a+b)(a^2+ab+b^2); a^2-b^2 = (a+b)(a-b); in case of cubic and bi-quadratic polynomials we can find linear factor by checking the sum of the coefficients, if the sum of the coefficients is zero then x = 1 is a solution and (x-1) a factor. Lastly if the sum of the coefficients of the even powers of the variable is equal to the sum of the coefficients of the odd powers of the variables then x = -1 is a solution and (x+1) is a factor.

**Soroban** · February 6th 2013, 07:47 PM

Hello, skg94!

You are expected to know basic factoring and these three formulas:

. . Difference of squares: . $a^2 - b^2 \:=\:(a-b)(a+b)$

. . Sum/difference of cubes: . $a^3 \pm b^3 \:=\:(a\pm b)(a^2 \mp ab + b^2)$

$[1]\;\frac{\frac{1}{\sqrt{x}} - 1}{x-1}$

$\frac{\frac{1}{\sqrt{x}} - 1}{x-1} \;=\;\frac{\frac{1-\sqrt{x}}{\sqrt{x}}}{x-1} \;=\; \frac{1-\sqrt{x}}{\sqrt{x}(x-1)} \;=\;-\frac{\sqrt{x}-1}{\sqrt{x}(x-1)}$

Here's where it gets sneaky . . . $x-1$ is a difference of squares.

. . $x - 1 \;=\;(\sqrt{x})^2 - 1^2 \;=\;(\sqrt{x} - 1)(\sqrt{x} + 1)$

The fraction becomes: . $-\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)} \;=\;-\frac{1}{\sqrt{x}(\sqrt{x}+1)}$

$[2]\; \frac{x\sqrt{x} - 8}{x-4}$

We have: . $\frac{\overbrace{(\sqrt{x})^3 - 2^3}^{\text{diff.cubes}}}{\underbrace{(\sqrt{x})^2 -2^2}_{\text{diff.squares}}} \;=\;\frac{(\sqrt{x}-2)(x + 2\sqrt{x}+4)}{(\sqrt{x}-2)(\sqrt{x}+2)} \;=\;\frac{x+2\sqrt{x}+4}{\sqrt{x}+2}$

$(a)\;x^6+8$ . sum of cubes

$(x^2)^3 + 2^3 \;=\;(x^2+2)(x^4 - 2x^2 + 4)$

$(b)\;x^4-16$ . diff. of squares

$(x^2)^2 - 4^2 \;=\;\overbrace{(x^2 - 4)}^{{\color{blue}\text{diff.squares}}}(x^2+4) \;=\;(x-2)(x+2)(x^2+4)$

$(c)\;r^8-1$ .diff. of squares

$(r^4)^2 - 1^2 \;=\;\overbrace{(r^4-1)}^{{\color{blue}\text{diff.squares}}}(r^4+1) \;=\;\overbrace{(r^2-1)}^{{\color{blue}\text{diff.squares}}}(r^2+1)(r^4 +1)$

. . . . . . . . . . $=\;(r-1)(r+1)(r^2+1)(r^4+1)$

$(d)\;x^3-x^2-16x+16$

$x^3-x^2-16x + 16 \;=\;x^2(x-1) - 16(x-1)$

. . . . . . $=\;(x-1)(x^2-16) \;=\;(x-1)(x-4)(x+4)$

$(e)\;x^4-3x^3-7x^2+27x-18$

We have: . $x^4 - 3x^3 - 7x^2 + 27x -18$

. . $=\; x^4-3x^3 + \overbrace{2x^2 - 9x^2} + 27x - 18$

. . $=\;x^2(x^2-3x+2) - 9(x^2-3x+2)$

. . $=\;(x^2-3x+2)(x^2-9)$

. . $=\'(x-1)(x-2)(x-3)(x+3)$

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