Important Questions: Polynomials - Class 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 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