3/(x-3)+(2x+8)/(x^2-9)*(x^2+3x)/(4x-16)

Question

Year 11, Preliminary HSC question. Operations with algebraic fractions. Please simplify the question at hand.

Answer 1

Note: I use * for multiply, ^ for exponent, and <> for not equal.

 

3/(x-3) + (2x+8)/(x^2-9) * (x^2+3x)/(4x-16)

3/(x-3) + 2(x+4)/((x+3)(x-3)) * x(x+3)/(4(x-4))

Can’t divide by zero means x <> -3, 3, 4.

Do multiplication of fractions:

3/(x-3) + 2x(x+3)(x+4)/(4(x-4)(x+3)(x-3)), x <> -3, 3, 4.

3/(x-3) + x(x+3)(x+4)/(2(x-4)(x+3)(x-3)), x <> -3, 3, 4.

(x+3)/(x+3) = 1 except when x = -3 it’s undefined.
=> there’s a “hole” at x = -3.

Now that we’ve noted and excluded x = -3 we can cancel the factors:

3/(x-3) + x(x+4)/(2(x-4)(x-3)), x <> -3, 3, 4.

Get common denominator:

3*2(x-4)/(2(x-4)(x-3)) + x(x+4)/(2(x-4)(x-3)), x <> -3, 3, 4.

Add fractions:

(6(x-4) + x(x+4))/(2(x-4)(x-3)), x <> -3, 3, 4.

(6x - 24 + x^2 + 4x)/(2(x-4)(x-3)), x <> -3, 3, 4.

(x^2 + 10x - 24)/(2(x-4)(x-3)), x <> -3, 3, 4.

(x-2)(x+12)/(2(x-4)(x-3)), x <> -3, 3, 4.

If you were graphing this as a function, it would have:

A hole at x = -3;
Vertical asymptotes at x = 3, 4;
Horizontal asymptote at y = 1/2 because x^2/(2x^2) = 1/2;
Zeros at x = -12, 2; b/c numerator zero there;
y-intercept = (-2)(12)/(2(-4)(-3)) = -1