Polynomials | Maths | Class 10th | Chapter 2 | PYQ Level 1

Polynomials :- PYQ :- Level 1

1. Find the value of “p” from the polynomial x2 + 3x + p, if one of the zeroes of the polynomial is 2.

Solution:

As 2 is the zero of the polynomial.

We know that if α is a zero of the polynomial p(x), then p(α) = 0

Substituting x = 2 in x2 + 3x + p,

⇒ 22 + 3(2) + p = 0

⇒ 4 + 6 + p = 0

⇒ 10 + p = 0

⇒ p = -10

2. Does the polynomial a4 + 4a2 + 5 have real zeroes?

Solution:

In the aforementioned polynomial, let a2 = x.

Now, the polynomial becomes,

x2 + 4x + 5

Comparing with ax2 + bx + c,

Here, b2 – 4ac = 42 – 4(1)(5) = 16 – 20 = -4

So, D = b2 – 4ac < 0

As the discriminant (D) is negative, the given polynomial does not have real roots or zeroes.

3. Compute the zeroes of the polynomial 4x2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.

Solution:

Let the given polynomial be p(x) = 4x2 – 4x – 8

To find the zeroes, take p(x) = 0

Now, factorise the equation 4x2 – 4x – 8 = 0

4x2 – 4x – 8 = 0

4(x2 – x – 2) = 0

x2 – x – 2 = 0

x2 – 2x + x – 2 = 0

x(x – 2) + 1(x – 2) = 0

(x – 2)(x + 1) = 0

x = 2, x = -1

So, the roots of 4x2 – 4x – 8 are -1 and 2.

Relation between the sum of zeroes and coefficients:

-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x2)

Relation between the product of zeroes and coefficients:

(-1) × 2 = -2 =  -8/4 i.e (constant/coefficient of x2)

4. Find the quadratic polynomial if its zeroes are 0, √5.

Solution:

A quadratic polynomial can be written using the sum and product of its zeroes is:

x2 – (α + β)x + αβ

Where α and β are the roots of the polynomial.

Here, α = 0 and β = √5

So, the polynomial will be:

x2 – (0 + √5)x + 0(√5)

⇒ x2 – √5x

5. Find the value of “x” in the polynomial 2a2 + 2xa + 5a + 10 if (a + x) is one of its factors.

Solution:

Say, f(a) = 2a2 + 2xa + 5a + 10

Since, (a + x) is a factor of 2a2 + 2xa + 5a + 10, f(-x) = 0

So, f(-x) = 2x2 – 2x2 – 5x + 10 = 0

Or, -5x + 10 = 0

Thus, x = 2

6. How many zeros does the polynomial (x – 3)2 – 4 can have? Also, find its zeroes.

Solution:

Given equation is (x – 3)2 – 4

Now, expand this equation.

=> x2 + 9 – 6x – 4

= x2 – 6x + 5

As the equation has a degree of 2, the number of zeroes will be 2.

Now, solve x2 – 6x + 5 = 0 to get the roots.

So, x2 – x – 5x + 5 = 0

=> x(x – 1) -5(x – 1) = 0

=> (x – 1)(x – 5) = 0

x = 1, x = 5

So, the roots are 1 and 5.

7. α and β are zeroes of the quadratic polynomial x2 – 6x + y. Find the value of ‘y’ if 3α + 2β = 20.

Solution:

Let, f(x) = x² – 6x + y

From the given,

3α + 2β = 20———————(i)

From f(x),

α + β = 6———————(ii)

And,

αβ = y———————(iii)

Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),

=> α = 20 – 12 = 8

Now, substitute this value in equation (ii),

=> β = 6 – 8 = -2

put the value of α and β in equation (iii) to get the value of y, such as;

y = αβ = (8)(-2) = -16

8.  If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, then find the value of a and b.

Solution:

Let the given polynomial be:

p(x) = x3 – 3x2 + x + 1

Given,

The zeroes of the p(x) are a – b, a, and a + b.

Now, compare the given polynomial equation with general expression.

px3 + qx2 + rx + s = x3 – 3x2 + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

For sum of zeroes:

Sum of zeroes will be = a – b + a + a + b

-q/p = 3a

Substitute the values q and p.

-(-3)/1 = 3a

Or, a = 1

So, the zeroes are 1 – b, 1, 1 + b.

For the product of zeroes:

Product of zeroes = 1(1 – b)(1 + b)

-s/p = 1 – 𝑏2

=> -1/1 = 1 – 𝑏2

Or, 𝑏2 = 1 + 1 =2

So, b = √2

Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 𝑥3 − 3𝑥2 + 𝑥 + 1.

9. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.

(i) 1/4, -1

(ii) 1,1

(iii) 4, 1

Solution:

(i) From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β
Product of zeroes = αβ

Given,

Sum of zeroes = 1/4
Product of zeroes = -1

Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ

x2 – (1/4)x + (-1)

4x2 – x – 4

Thus, 4x2 – x – 4 is the required quadratic polynomial.

(ii) Given,
Sum of zeroes = 1 = α + β
Product of zeroes = 1 = αβ

Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x2 – (α + β)x + αβ

x2 – x + 1

Thus, x2 – x + 1 is the quadratic polynomial.

(iii) Given,
Sum of zeroes, α + β = 4
Product of zeroes, αβ = 1

Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x2 – (α + β)x + αβ

x2 – 4x + 1

Thus, x2 – 4x +1 is the quadratic polynomial.

10. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and-√(5/3).

Solution: Since this is a polynomial of degree 4, hence there will be a total of 4 roots.

√(5/3) and-√(5/3) are zeroes of polynomial f(x).
∴ [x-√(5/3)] [x+√(5/3)] = x2-(5/3)

Therefore, 3x2 + 6x + 3 = 3x(x + 1) +3 (x + 1)
= (3x + 3)(x + 1)
= 3(x + 1)(x + 1)
= 3(x + 1)(x + 1)
Hence, x + 1 = 0 i.e. x = – 1 , – 1 is a zero of p(x).
So, its zeroes are given by: x = −1 and x = −1.
Therefore, all four zeroes of the given polynomial are:
√(5/3) and-√(5/3), −1 and −1.

I hope this helps you in excelling in your academics. Please check out the Level 2 PYQs of this chapter. If you have any queries, please do let me know in the comments section.


Thanking You.