It is given that the polynomial P(x) = x3 ax2 bx c has three dis...
Given Information:
Polynomial P(x) = x3 ax2 bx c has three distinct positive integer roots and P(22) 21. Q(x) = x2- 2x 22. P(Q(x)) has no real roots.
To Find:
Value of a in P(x).
Solution:
Let the roots of P(x) be p, q, and r. Then we have:
- p + q + r = a
- pqr = -c
- p + q + r = a
- pq + qr + rp = b
Since P(x) has three distinct positive integer roots, we can assume that p < q="" />< r="" are="" positive="" integers.="" we="" are="" also="" given="" that="" p(22)="21," which="" means="" />
P(22) = (22 - p)(22 - q)(22 - r) = 21
Since 22 - p, 22 - q, and 22 - r are positive integers, we can only have:
22 - p = 1, 3, or 7
22 - q = 1, 3, or 7
22 - r = 1, 3, or 7
Without loss of generality, we can assume that:
22 - p = 1, 22 - q = 3, and 22 - r = 7
which gives us:
p = 21, q = 19, and r = 15
Now, let us consider P(Q(x)):
P(Q(x)) = (Q(x) - p)(Q(x) - q)(Q(x) - r)
Substituting Q(x) = x2 - 2x + 22, we get:
P(Q(x)) = (x2 - 2x + 22 - p)(x2 - 2x + 22 - q)(x2 - 2x + 22 - r)
Expanding the above expression and simplifying, we get:
P(Q(x)) = x6 - (p + q + r + 6)x4 + (pq + qr + rp + 8(p + q + r) + 44)x2 - pqr - 44(p + q + r) - 484
Substituting the values of p, q, and r, we get:
P(Q(x)) = x6 - 72x4 + 1330x2 - 13515
Since P(Q(x)) has no real roots, its discriminant is negative:
72^2 - 4(1330) < />
which gives us:
a = p + q + r = 21 + 19 + 15 = 55
Answer: