CAT — Polynomials

Given, 2|x|(x² + 1) = 5x²
Since the equation contains |x|, consider two cases.
Case 1: x ≥ 0
Then, |x| = x
So, 2x(x² + 1) = 5x² → 2x³ + 2x = 5x² → 2x³ − 5x² + 2x = 0
x(2x² − 5x + 2) = 0
Factorize, x(2x − 1)(x − 2) = 0
Possible solutions are, x = 0, 1/2, 2
Among these, the integer solutions are, x = 0, 2
Case 2: x < 0
Then, |x| = −x
So, −2x(x² + 1) = 5x² → 2x³ + 5x² + 2x = 0
x(2x² + 5x + 2) = 0
Factorize, x(2x + 1)(x + 2) = 0
Possible solutions are, x = 0, −1/2, −2
Since x < 0, discard x = 0.
The only integer solution is, x = −2
Therefore, the integer solutions are, x = −2, 0, 2
Hence, the number of integer solutions is 3
Answer: 3

Given, x³ + (2r + 1)x² + (4r − 1)x + 2 = 0
One of the roots is −2.
Substitute x = −2.
(−2)³ + (2r + 1)(−2)² + (4r − 1)(−2) + 2 = 0
−8 + 4(2r + 1) − 2(4r − 1) + 2 = 0
−8 + 8r + 4 − 8r + 2 + 2 = 0
0 = 0
Hence, −2 is a root for every value of r.
Now divide the cubic by (x + 2).
Using synthetic division with −2:
-2 | 1 2r+1 4r−1 2
-2 -4r+2 -2
1 2r−1 1 0
So the remaining quadratic is x² + (2r − 1)x + 1 = 0
For the other two roots to be real, the discriminant must be non-negative.
Discriminant = (2r − 1)² − 4 = 4r² − 4r + 1 − 4 = 4r² − 4r − 3
Therefore, 4r² − 4r − 3 ≥ 0
Now test the non-negative integers.
For r = 0: 4(0)² − 4(0) − 3 = −3 < 0 → Not possible.
For r = 1: 4(1)² − 4(1) − 3 = −3 < 0 → Not possible.
For r = 2: 4(2)² − 4(2) − 3 = 16 − 8 − 3 = 5 > 0 → Possible.
Hence, the minimum non-negative integer value of r is 2
Answer: 2