Math Olympiad Test: Polynomials - 3 - Free MCQ with solutions Class 10

Sum of zeroes of polynomial
5x² – (3 + k)x + 7 is 
According to the question,
((3 + k)/5) = 0 ⇒ k = –3
Now, 2x² – 2(k + 11)x + 30 becomes 2x² – 16x + 30.
i.e., 2x² – 16x + 30 = 0 or x² – 8x + 15 = 0
⇒ x = 3, 5
Hence, zeroes of polynomial 2x² – 16x + 30 are 3, 5.

For a cubic polynomial to exist, coefficient of term x³ must not be equal to zero.

Given equation is x² + px + 12 = 0
Now, if a and b are its roots, then sum of roots, a + b = –p and product of roots, a × b = 12
Also, a – b = 1 (Given)
We know that, (a – b)² = (a + b)² – 4ab
⇒ 1 = p² – 4 × 12 ⇒ 1 = p² – 48 ⇒ p² = 49 ⇒ p = ±7

Given polynomial is 2x³ – 3x² + 4tx – 5
Sum of product of roots taken two at a time is 4t/2.
∴ 4t/2 = -8 ⇒ t = –4

For a cubic polynomial, ax³ + bx² + cx + d
Sum of zeroes = -(b/a)
Sum of the product of zeroes taken two at a time = (c/a).
Product of zeroes = -(d/a)
We have, -(b/a) = -3, (c/a) = 8 and (-d)/a = 4
∴ x³ + 3x² + 8x – 4 is the required polynomial.

Since, 3 is one of the zeroes of polynomial p(x).
So, p(3) = 0

This can be done either by long division as shown in the image
We can see the division gives us a quotient (x − 2) and a remainder 3

Given equation is x² + k(x – 1) – c
= x² + kx – (k + c)
Since, p and q are the zeroes,
∴ = – k and pq = – (k + c)
Now, (p – 1) (q – 1) = pq – q – p + 1
= pq – (p + q) + 1 = – (k + c) – (–k) + 1
= – k – c + k + 1 = 1 – c

For more than three distinct real roots the graph must cut x-axis at least four times. So, graph in option (C) has more than three distinct real roots.

By long division method, we have

We must subtract the remainder so that f(x) is exactly divisible by x² + x – 1
Hence, –38x – 60 is to be subtracted.