Just factor right?

Algebra Level 1

x 3 + 2 x 2 + x + 2 x 2 − 4 = 0 \large \frac{x^3 + 2x^2 + x + 2}{x^2- 4} = 0

What real value of x x satisfies the equation above?

2 -1 No solution -2

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Harshit Singhania by Harshit Singhania , 33, India

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1 solution

Aeshit Singh
Jun 13, 2015

Factoring the numerator gives us x 3 + 2 x 2 + x + 2 = ( x + 2 ) ( x 2 + 1 ) . x^{3}+2x^2+x+2 = (x+2)(x^2+1).

The fraction is now: ( x + 2 ) ( x 2 + 1 ) ( x + 2 ) ( x − 2 ) \frac{(x+2)(x^2+1)}{(x+2)(x-2)} From here, we see the factor of ( x + 2 ) (x+2) can cancel, but we note that x = − 2 x = -2 is not a possible solution because it causes the original expression to be undefined. However, the numerator is a quadratic expression which is positive for all real values of x x . Therefore there are no real solutions.

3 Helpful 1 Interesting 0 Brilliant 0 Confused

Moderator note:

Nicely done.

Bonus question : Can you determine the asymptote for the function f ( x ) = x 3 + 2 x 2 + x + 2 x 2 − 4 f(x) = \frac{x^3+2x^2+x+2}{x^2-4} ?

Thank You, Challenge Master

Now, for the bonus question

By long division method, x 3 + 2 x 2 + x + 2 = ( x 2 − 4 ) ( x + 2 ) + ( 5 x + 6 ) x^{3}+2x^2+x+2=(x^{2}-4)(x+2) +(5x+6) The quotient becomes the slant asymptote: y=x+2

Aeshit Singh - 6 years ago

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there are more asymptotes for the same function. that is x=2 and x=-2 are vertical asymptotes and that y = 2 must be a horizontal asymptote. sorry for incomplete solutions,i am having a lot of troubles in understanding formatting guide

Mohammed owais Khokhar - 5 years, 12 months ago

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