Question

Find the illegal values of c in the multiplication statement. c²-3c-10/c²+5c-14 · c²-c-2/c²-2c-15

A. c = –7, c = –3, c = –2, and c = 5
B. c = –7, c = –3
C. c = 7, c = 3, c = –2, and c = –5
D. c = –7, c = –3, c = 2, and c = 5

Answer 1

The illegal values of c in the multiplication statement
$\frac{c^2-3c-10}{c^2+5c-14}\cdot \frac{c^2-c-2}{c^2-2c-15}$
are the values of c for which
$c^2+5c-14=0$ and $c^2-2c-15=0$

For
$c^2+5c-14=0 \\ \\ (c-2)(c+7)=0 \\ \\ c-2=0 \ and \ c+7=0 \\ \\ c=2 \ and \ c=-7$

For
$c^2-2c-15=0 \\ \\ (c-5)(c+3)=0 \\ \\ c-5=0 \ and \ c+3=0 \\ \\ c=5 \ and \ c=-3$

Therefore, the illegal values of c in the multiplication statement
$\frac{c^2-3c-10}{c^2+5c-14}\cdot \frac{c^2-c-2}{c^2-2c-15}$
are -7, -3, 2 and 5.

Answer 2

Answer: D. PLZ MARK BRAINLIEST THIS TOOK ME ABOUT 5 MIN TO DO

2,-7,5,and -3

Step-by-step explanation:

so look at the denominators... the illegal value means that you cant practically have a zero in your answer (so we are trying to find what values = 0)

You have 2 denominators which are c2 +5c-14=0 and c2 -2c -15=0.

1) Factor out c2 +5c-14=0 ... (c-2)(c+7)=0

2) Factor out c2 -2c -15=0 ... (c-5)(c+3)=0

c-2=0 {2} c+7=0 {-7} (2 and -7 are the illegal values for c2 +5c-14=0 )

c-5=0 {5} c+3=0 {-3} (5 and -3 are the illegal values for c2 -2c -15=0 )

yep so your illegal values are 2,-7,5, and  -3