Question

how to find the x-intercepts of (5x-3)(x+7)/(-3x-1)(7x-3)

Answer 1

Hm let's see
Assuming I'm reading this right:
\[
f(x) = \frac{(5x-3)(x+7)}{(-3x-1)(7x-3)}
\]
Well first, we know that the divisor can't be 0.. so
\[
(-3x-1)(7x-3) \ne 0 \quad \text{ means:}\\
-3x-1\ne 0 \quad \text{and} \quad 7x-3 \ne 0 \\
x \ne -\frac{1}{3} \quad \text{and} \quad x \ne \frac{3}{7}
\]
Now we say:
\[
f(x) = 0 \quad \implies \quad \frac{(5x-3)(x+7)}{(-3x-1)(7x-3)} = 0
\]
For f(x) to be 0 the numerator has to be 0.. (or simply multiply both side by the divisor..)
\[
(5x-3)(x+7) = 0 \quad \text{ means:}\\
5x-3 = 0 \quad \text{or} \quad x+7 = 0 \\
x = \frac{3}{5} \quad \text{or} \quad x = -7 \\
\]
Since both results are legal (won't make a division by 0), then we can say, the function has value 0 (intersects X axis) when x =3/5 or x = -7

Answer 2

thank you!