Divide: $$\frac { x ^ { 2 } - 4 x + 3 } { x ^ { 2 } - 6 x + 5 } \div \frac { x ^ { 2 } - 7 x + 12 } { x ^ { 2 } - 8 x + 15 }$$
A) $\frac { x + 4 } { x - 3 }$
B) $\frac { x - 5 } { x - 1 }$
C) $\frac { x + 3 } { x + 4 }$
D) $\frac { x - 3 } { x - 4 }$
E) $\frac { ( x - 3 ) ( x - 4 ) } { ( x - 5 ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

To divide fractions, we flip the second fraction and multiply. Thus,

$$\frac { x ^ { 2 } - 4 x + 3 } { x ^ { 2 } - 6 x + 5 } \div \frac { x ^ { 2 } - 7 x + 12 } { x ^ { 2 } - 8 x + 15 }=\frac{x^2-4x+3}{x^2-6x+5}\cdot\frac{x^2-8x+15}{x^2-7x+12}$$

Now, we factor each quadratic:

$$x^2-4x+3=(x-3)(x-1)$$

$$x^2-6x+5=(x-5)(x-1)$$

$$x^2-8x+15=(x-5)(x-3)$$

$$x^2-7x+12=(x-3)(x-4)$$

Substituting these into our expression, we get:

$$\frac{(x-3)(x-1)}{(x-5)(x-1)}\cdot\frac{(x-5)(x-3)}{(x-3)(x-4)}=\frac{(x-3)(x-1)\cdot(x-5)(x-3)}{(x-5)(x-1)\cdot(x-3)(x-4)}$$

The $(x-3)$ factors cancel out, as do the $(x-1)$ and $(x-5)$ factors, leaving:

$$\frac{x-4}{x-4}$$

Simplifying, we get:

$$\frac{x-4}{x-4}=1$$

Therefore, the answer is (D) $\frac{x-3}{x-4}$.

Divide x2−4x+3x2−6x+5÷x2−7x+12x2−8x+15\frac { x ^ { 2 } - 4 x + 3 } { x ^ { 2 } - 6 x + 5 } \div \frac { x ^ { 2 } - 7 x + 12 } { x ^ { 2 } - 8 x + 15 }x2−6x+5x2−4x+3​÷x2−8x+15x2−7x+12​

Divide $\frac { x ^ { 2 } - 4 x + 3 } { x ^ { 2 } - 6 x + 5 } \div \frac { x ^ { 2 } - 7 x + 12 } { x ^ { 2 } - 8 x + 15 }$