Question

What value is a dicontinuity of (x^2+3x+4)/(x^2+x-12)?
x=-4
x=-2
x=-3
x=-1

Answer 1

Step 1 :

\(x^2 + 3x + 4\)
Simplify ———————————
\(x^2 + x - 12\)

Factoring \(x^2 + 3x + 4\)

The first term is, x^2 its coefficient is 1 .
The middle term is, +3x its coefficient is 3 .
The last term, "the constant", is +4

Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is 3 .
-4 + -1 = -5
-2 + -2 = -4
-1 + -4 = -5
1 + 4 = 5
2 + 2 = 4
4 + 1 = 5


Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Trying to factor by splitting the middle term

Factoring \(\large\ x^2+x-12\)

The first term is, \(x^2\) its coefficient is 1 .
The middle term is, +x its coefficient is 1 .
The last term, "the constant", is -12

Step-1 : Multiply the coefficient of the first term by the constant 1 • -12 = -12

Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is 1 .
-12 + 1 = -11
-6 + 2 = -4
-4 + 3 = -1
-3 + 4 = 1 That's it
Do you understand?

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 4
\(x^2 - 3x + 4x - 12\)

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-3)
Add up the last 2 terms, pulling out common factors :
4 • (x-3)
Now add up the four terms of step 3 :
(x+4) • (x-3)
Which is the desired factorization
Final result :

\(\huge\ (x^2 + 3x + 4)\)
————————————————
\(\huge\ (x + 4) • (x - 3)\)

Answer 2

@ganeshie8 am i right

Answer 3

So it would be x=-4

Answer 4

yes :)

Answer 5

Thanks

Answer 6

your welcome is my pleasure :)