Important Questions: Polynomials - Grade 10 MCQ

Sum of zeros = 3/1
-b/a = 3/1
Product of zeros = 2/1
c/a = 2/1
This gives 
a = 1
b = -3
c = -2,
The required quadratic equation is
ax²+bx+c
So,  x²-3x+2

Using mid-term splitting,
x² + x - 20 = x² + 5x - 4x - 20 = x(x + 5) -4(x + 5)
Taking common x + 5
(x + 5)(x - 4), so the other factor is x - 4

P(x) = 2x² + 5x + 1
Sum of roots = -5/2
Product of roots = 1/2
Therefore substituting these values, 
α + β +αβ 
=(α + β) + αβ
= -5/2 + 1/2
= -4/2 
= -2

If α  and β are the zeros of the polynomial then
(x−α)(x−β) are the factors of the polynomial
Thus, (x−α)(x−β) is the polynomial.
So, the polynomial =x² − αx − βx + αβ
=x² − (α + β)x + αβ....(i) 
Now,the quadratic polynomial is  
2 − 3x − x² = x² + 3x − 2....(ii)

Now, comparing equation (i) and (ii),we get,
−(α + β) = 3 
α + β = −3

The quadratic equation is of the form  x² - (sum of zeros) x + (product of zeros)
=x² - 5x - 14

Zeros of the polynomials are the values which gives zero when their value is substituted in the polynomial
When x = 2,
x² + ax + b = (2)² + a*2 + b = 0
4 + 2a + b = 0
b = - 4 - 2a    ….1
When x = 3,
(3)² + 3a + b = 0
9 + 3a + b = 0
Substituting 
9 + 3a - 4 - 2a = 0
5 + a = 0
a = - 5
b = 6

Given:
• The sum of the zeros is − 5
• The product of the zeros is 3
For a quadratic polynomial x² + bx + c (where a = 1), we can use these relationships:
• The sum of the zeros - b/1 = - 5 gives b = 5
• The product of the zeros c/1 = 3 gives c = 3
Therefore, the quadratic polynomial is:
x² + 5x + 3

Given α and β are the zeroes of the polynomial x² − 5x + k
Also given that α − β = 1 → (1)
Recall that sum of roots (α + β) = −(b/a)
∴ α + β = 5 → (2)
Add (1) and (2), we get
α − β = 1
α + β = 5
2α = 6
∴ α = 3
Put α = 3 in α + β = 5
3 + β = 5
∴ β = 2
Hence 3 and 2 are zeroes of the given polynomial
Put x = 2 in the given polynomial to find the value of k ( Since 2 is a zero of the polynomial, f(2) will be 0 )
x² − 5x + k = 0
⇒ 2² − 5(2) + k = 0
⇒ 4 − 10 + k = 0
⇒ − 6 + k = 0
∴ k = 6

p(x) = ax² + bx + c
A quadratic polynomial, by definition, is a degree 2 polynomial, meaning it can have at most 2 distinct zeros (solutions).
The number of zeros (solutions) of the polynomial depends on the discriminant Δ = b² - 4ac:
• If Δ > 0, the quadratic has two distinct real zeros.
• If Δ = 0, the quadratic has exactly one real zero (a repeated root).
• If Δ < 0, the quadratic has no real zeros but two complex zeros.
Thus, a quadratic polynomial can have at most two zeros.

Given that the sum of the zeros is 2 and the product is 4, the quadratic equation has the form:
f(x) = a(x² - 2x + 4)
Setting a = 2 (to match the coefficients), we get:
f(x) = 2(x² - 2x + 4) = 2x² - 4x - 1
So, the correct answer is:
d) 2x² - 4x - 1.