When dividing rational expressions, how do you know what x CANT BE? HELLP FINALS TOMORROW!!!?

Question

When dividing rational expressions, how do you know what x CANT BE? HELLP FINALS TOMORROW!!!?
okay for example,

x^2+5x+10/x+6 DIVIDED BY x^2+6x-36/ x-2


okay so, i know that i flip the second fraction so the two fractions are multiplied, but BEFORE I FLIPP IT, is the x-2 on the denominator supposed to be like x cant be 2..? or do you only do the "x cant be _" after you flip the second fraction?

Answer 1

(x^2 + 5x + 10)/(x + 6) divided by (x^2 + 6x - 36)/(x - 2)

x ≠ 2, -6
because then both the rational expression in the original question will cease to be defined.

If x is 2, or -6, then you are dividing a defined quantity with an undefined quantity. Can you do that?

Now after flipping, x^2 + 6x - 36 is the denominator of the second expression. The first is unchanged

Now x^2 + 6x - 36 ≠ 0 (because you can't multiply a defined quantity by an undefined one!)

x ≠ [- 6 +/- sqrt (36 + 144)]/2
x ≠ (- 6 +/- sqrt 180)/2
x ≠ (-6 +/- 6 sqrt 5)/2
x ≠ -6 (1 +/- sqrt 5)/2
x ≠ -3 +/- 3 sqrt 5



So,
x ≠ -6, 2, -3 + 3 sqrt 5, -3 - 3 sqrt 5

Total 4 number x cannot equal

Answer 2

For most expressions, we assume that we are operating in the domain of the expression.

Division by zero is not defined. Division by zero occurs when:

x + 6 = 0
x = -6

x^2 + 6x - 36 = 0
x = -3 + 3sqrt(5) or x = -3 - 3sqrt(5)

x - 2 = 0
x = 2

So, we assume for this expression that x is any real number except -6, 2, -3 + 3sqrt(5), and -3 - 3sqrt(5).

However, this is not an equation to be solved, where we must identify extraneous roots. This is a division problem involving polynomials. So, multiply by the reciprocal of the divisor (flip it) and factor and simplify.

x^2 + 5x + 10..............x - 2
------------------- * -------------------
x + 6.....................x^2 + 6x - 36

STOP! This is the ANSWER. The numerators and denominators can not be factored. This is as far as we should go.

Answer 3

Upon dividing 1st expression by 2nd expression:-
[(x² + 5x + 10) / (x + 6)] X [(x - 2) / (x² + 6x - 36)]
[(x² + 5x + 10).(x - 2)] / [(x + 6).(x² + 6x - 36]
Denominator cannot = 0, thus:-
x ≠ - 6 and (x² + 6x - 36) ≠ 0
IF x² + 6x - 36 = 0
x = [- 6 ± √(36 + 144)] / 2
x = [- 6 ± √180 ] / 2
x = [- 6 ± 6√5] / 2
x = -3 ± 3√5
But denominator CANNOT = 0 therefore:-
x ≠ (-3 ± 3√5)
x ≠ 6 (from above)

Answer 4

1) (x^2 - 12x + 32)/(x^2 - 6x - 16) = (x^2 - 4x - 8x + 32)/(x^2 + 2x - 8x - 16) = (x - 4)(x - 8)/(x + 2)(x - 8) (cancel out x - 8) = (x - 4)/(x + 2) 2) (x^2 - x - 12)/(x^2 - 5x - 24) = (x^2 + 3x - 4x - 12)/(x^2 + 3x - 8x - 24) = (x + 3)(x - 4)/(x + 3)(x - 8) (cancel out x + 3) = (x - 4)/(x - 8)

Answer 5

in the example you've given..

x can't be neither 2 nor -6...
if u do it before flippin.

i think you must think of the "x can't be" thing after flippin.
if you don't flip it and say x can't be 2(like in the above ex) it is not wrong cause you've got a valid reason.