find the missing integer 1/(x+3) + 1/(x+?) + 1/(x+?) = 3x^2+12x-13 / (x-3)(x+2)(x+7)
I think you’ve made a mistake in your first fraction, 1/(x+3). I believe that should be 1/(x – 3).
Your lhs should be 1/(x-3) + 1/(x+A) + 1/(x+B) =
I’ve used A and B here to show that there are two distinct values for the question mark (?).
Now give the partial fractions a common denominator to match up with the denominator of the ration function on the rhs. This comes out as,
(x+A)(x+B) / [(x-3)(x+A)(x+B)] + (x-3)(x+B) / [(x-3)(x+A)(x+B)] + (x-3)(x+A) / [(x-3)(x+A)(x+B)] =
Taking out the common denominator gives us,
[(x+A)(x+B) + (x-3)(x+B) + (x-3)(x+A)] / [(x-3)(x+A)(x+B)] =
Now insert the rational function on the rhs of teh above expression.
[(x+A)(x+B) + (x-3)(x+B) + (x-3)(x+A)] / [(x-3)(x+A)(x+B)] = 3x^2+12x-13 / (x-3)(x+2)(x+7)
We see that, when multiplied out, the numerator on the lhs will simply be some quadratic expression or other. The lhs will be a quadratic expression over its denominator. The rhs is also a quadratic expression over its denominator. Since they are similar in form, and each side is reduced to its lowest form, then we can equate the numerators and denominators of each side with each other. That means that
(x-3)(x+A)(x+B) = (x-3)(x+2)(x+7)
giving A = 2, B = 7
And
(x+A)(x+B) + (x-3)(x+B) + (x-3)(x+A) = 3x^2+12x-13
Substituting for A =2 and B = 7,
(x+2)(x+7) + (x-3)(x+7) + (x-3)(x+2) = 3x^2+12x-13
Multiplying out the lhs,
x^2 + 9x + 14 + x^2 + 4x – 21 + x^2 – x – 6 = 3x^2 + 12x – 13
3x^2 + 12x – 13 = 3x^2 + 12x – 13
Which confirms the values for A and B.
The two unknown values are: A = 2, B = 7