+24 Consider the two expressions $\frac{6x^3+9x^2+4x+7}{2x+3}$ and $3x^2+2+\frac1{2x+3}.$
a) Show that the two expressions represent equal numbers when $x=10.$
b) Explain why these two expressions do not represent equal numbers when $x=-\dfrac32.$
c) Show that these two expressions represent equal numbers for all $x$ other than $-\dfrac32.$
+1632 \({6x^3+9x^2+4x+7\over{2x+3}}, 3x^2 + 2 + {1\over{2x+3}}\)
a) To show that the two expressions represent equal numbers when x = 10, substitute 10 for x into each expression:
(6 * 10^3 + 9 * 10^2 + 4 * 10 + 7)/(23) = (6000 + 900 + 40 + 7)/23 = 6947/23
3 * 10^2 + 2 + 1/23 = 302 + 1/23 = 302*23/23 + 1/23, add numerators only: 6946/23 + 1/23 = 6947/23, equivalent to the expression above.
+1632 b) Explain why these two expressions do not represent equal numbers when \(x = -{3\over{2}}\).
b) You can show this by plugging -3/2 in- but wait! Can the denominator be 0? No, then the expression would be undefined!
The denominator, 2x + 3, is 2(-3/2)+3 = -3 + 3 = 0 when x = -3/2... So, these two expressions, each undefined, would not and could not be equal.
+1632 c) In part b we showed how these two expressions were undefined when x = -3/2, but now we must show why the two expressions are the same for all x except that one case- one can do this through manipulation...
The numerator, 6x^3 + 9x^2 + 4x + 7 can be manipulated and grouped to (2x + 3)(3x^2) + (2x + 3)(2) + 1, and you can see why they are the same after distribution *all I did was take the Greatest Common Factor = 2x + 3 out of the polynomial after grouping terms*
Then: (6x^3 + 9x^2 + 4x + 7)/(2x + 3) = [(2x + 3)(3x^2) + (2x + 3)(2) + 1]/(2x + 3). The 2x + 3 coefficients cancel out and you are left with a remainder of 1:
[(2x + 3)(3x^2) + (2x + 3)(2) + 1]/(2x + 3) = 3x^2 + 2 + 1/(2x + 3), which is the exact same as the second expression!
