Class 10 Maths Previous Year Questions - Polynomials

Ans: (b)
Let, f(x) = x² – 5x + 4 
Let p should be added to f(x) then 3 becomes zero of polynomial.
So, f(3) + p = 0 
⇒ (3)² – 5 × (3) + 4 + p = 0 
⇒ 9 + 4 – 15 + p = 0 
⇒ – 2 + p = 0 
⇒ p = 2 

So, 2 should be added.

Ans:
x² – 15 = 0
x² = 15
x = ± √15
Zeroes will be  α = √15 , β = – √15
Verification: Given polynomial is x² – 15
On comparing above polynomial with
ax² + bx + c, we have
a = 1, b = 0, c = –15
sum of zeros = α + β 

Product of zeros = αβ 

Hence, verified.

Ans: (b)
 Here, y = p(x) touches the x-axis at one point
So, number of zeros is one.

Ans: (a)

The polynomial is p(x) = x² + x - 1.

Step 1: The relationships between the zeroes and coefficients:

Sum of zeroes (α + β): - ba = - 11 = -1

Product of zeroes (αβ): ca = -11 = -1

Step 2: Simplify 1α + 1β:

1α + 1β = α + βαβ

Substitute the values:

α + βαβ = -1-1 = 1

Final Answer: (a) 1

Ans: (d)

The polynomial is p(x) = x² - 1.

Step 1:  Sum of zeroes (α + β): - ba = - 01

Step 2: Simplify:

- 01 = 0

Final Answer: (d) 0

Ans: (d) 

The polynomial is p(x) = 4x² - 3x - 7.

Step 1: calculating sum and product of zeroes 

Sum of zeroes (α + β): - ba = - (-3)4  = 34

Product of zeroes (αβ):  ca =  -74

Step 2: Simplify 1α + 1β:

α + βαβ  =  34-74  =  -37

Final Answer: (d) - 37

Ans: We have,

The polynomial is p(x) = 6x² + 37x - (k - 2).

Step 1: The relationship between the product of zeroes and coefficients:

Product of zeroes (αβ) =  ca = -(k - 2)6

It is given that αβ = 1. Substitute this:

-(k - 2)6 = 1

Step 2: Solve for k:

Multiply both sides by 6:

-(k - 2) = 6

Simplify:

k - 2 = -6

k = -4

Final Answer: k = - 4

Ans: (a)
 Given that -3 is a zero of quadratic polynomial (k - 1)²+ kx + 1.
⇒ Putting x = -3 in above equation, we get
∴ (k - 1) (-3)² + k(-3) +1 = 0
⇒ 9k - 9 - 3k + 1 = 0 ⇒ 6k - 8 = 0
⇒ k = 8/6
⇒ k = 4/3

Ans: (c)

The zeroes of the polynomial are -1 and 2.

Step 1: The polynomial with given zeroes is:

p(x) = a(x - α)(x - β)

Substitute the zeroes α = -1 and β = 2:

p(x) = a(x - (-1))(x - 2) = p(x) = a(x + 1)(x - 2)

Step 2: Expand the polynomial:

p(x) = a[(x)(x) + (x)(-2) + (1)(x) + (1)(-2)]

p(x) = a[x² - x - 2]

Step 3: Assuming a = 1:

p(x) = x² - x - 2

Final Answer: (c) x² - x - 2

Ans: (a)
 Given curve cuts the x-axis at four distinct points.
So, number of zeroes will be 4 .

Ans: (b)
The distance between point C and G is 6 units.

Ans: (a)
 Let α, β be the zeroes of required polynomial p(x).
Given, α + β=-5 and α.β=6
p(x) = x² - (Sum of zeros)x + (Product of zeros)
∴ p(x)=k[x² - (-5)x + 6] = k[x² + 5x + 6]  
Thus, one of the polynomial which satisfy the given condition is x²+ 5x + 6

Ans: Given, polynomial is f(x) =x² + 3x + k
Since, 2 is zero of the polynomial f(x).
∴ f(2) = 0
⇒ f(2) =(2)² + 3 x 2 + k
⇒  4 + 6 + k = 0
⇒ k = -10

Ans: (a)
 Since, the polynomial has two zeroes only. So. the degree of the polynomial is 2.

Ans: (b)
 Given, 2 is a zero of the polynomial
p(x) = x² + 3x + k
∴ p (2) = 0
⇒ (2)² + 3(2) + k = 0
⇒ 4 + 6 + k = 0 
⇒ 10 + k = 0
⇒ k= -10

Ans: (a)
 Let α, β be the zeroes of required polynomial p(x)
Given, α+ β = -5 and αβ = 6
p(x) = k[x² - (- 5)x + 6]
= k[x² + 5x + 6]
Thus, one of the polynomial which satisfy the given condition is x² + 5x + 6.

Ans: Let α, β be the zeroes of required polynomial Given, α + β = -3 and αβ = 2
∴ p(x) = k[x² - (-3)x + 2] = k(x² + 3x + 2)
For k = 1 , p (x) = x² + 3x + 2
Hence, one of the polynomial which satisfy the given condition is x² + 3x + 2.

Ans: (b)
Given:

$x^2\; -\; 3x\; -\; m(m\; +\; 3)\; =\; 0$x² − 3x − m(m + 3) = 0
Let's find the zeroes by applying the quadratic formula:

Substitute into the formula:

Simplify under the square root:

Taking the square root:

So, the zeroes are –m and m + 3.
Thus, the correct answer is (b) –m, m + 3.

Ans: 7
The given polynomial is x² -(k + 6)x + 2(2k - 1)
According to the question
Sum of zeroes = 1/2(Product of Zeroes ):
⇒ k + 6 = 1/2 x 2 (2k - 1)
⇒ k + 6 = 2k - 1
⇒ k = 7