RBSE Solutions Class 10 Maths Chapter 3 Polynomials Exercise 3.1

Polynomials Ex 3.1

Question 1. Find the zeros of the following quadratic polynomial and test the relation between zeros and coefficients.
(i) \( 4x^2 + 8x \)
(ii) \( 4x^2 - 4x + 1 \)
(iii) \( 6x^2 - x - 2 \)
(iv) \( x^2 - 15 \)
(v) \( x^2 - (\sqrt{3} + 1) x + \sqrt{3} \)
(vi) \( 3x^2 - x - 4 \)
Answer:
(i) For the polynomial \( f(x) = 4x^2 + 8x \):
First, factor the polynomial: \( f(x) = 4x(x + 2) \).
To find the zeros, set \( f(x) = 0 \):
\( 4x(x + 2) = 0 \)
This means either \( 4x = 0 \) or \( x + 2 = 0 \).
Solving these, we get \( x = 0 \) or \( x = -2 \).
So, the zeros are \( 0 \) and \( -2 \). Finding zeros helps us understand where the graph of the polynomial crosses the x-axis.

Now, let's check the relation between the zeros and coefficients:
Sum of zeros \( = 0 + (-2) = -2 \).
Using the formula \( -\frac{b}{a} \): for \( 4x^2 + 8x \), \( a=4, b=8, c=0 \).
\( -\frac{b}{a} = -\frac{8}{4} = -2 \). (The sum matches)
Product of zeros \( = 0 \times (-2) = 0 \).
Using the formula \( \frac{c}{a} \):
\( \frac{c}{a} = \frac{0}{4} = 0 \). (The product matches)
This confirms the relationship between the zeros and coefficients is correct.
In simple words: We find the values of \( x \) that make the polynomial zero. Then, we check if the sum and product of these zeros match the values we get from the formula using the numbers in the polynomial.

🎯 Exam Tip: Remember to clearly state the factors of the polynomial and then set each factor to zero to find the individual roots.

Question 1. (ii) \( 4x^2 - 4x + 1 \)
Answer:
For the polynomial \( f(x) = 4x^2 - 4x + 1 \):
This polynomial is a perfect square trinomial: \( f(x) = (2x - 1)^2 \).
To find the zeros, set \( f(x) = 0 \):
\( (2x - 1)^2 = 0 \)
\( 2x - 1 = 0 \)
\( 2x = 1 \)
\( x = \frac{1}{2} \)
So, the zeros are \( \frac{1}{2} \) and \( \frac{1}{2} \) (a repeated zero). A repeated zero means the graph touches the x-axis at that point but does not cross it.

Now, let's check the relation between the zeros and coefficients:
Sum of zeros \( = \frac{1}{2} + \frac{1}{2} = 1 \).
Using the formula \( -\frac{b}{a} \): for \( 4x^2 - 4x + 1 \), \( a=4, b=-4, c=1 \).
\( -\frac{b}{a} = -\frac{-4}{4} = \frac{4}{4} = 1 \). (The sum matches)
Product of zeros \( = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
Using the formula \( \frac{c}{a} \):
\( \frac{c}{a} = \frac{1}{4} \). (The product matches)
This confirms the relationship between the zeros and coefficients is correct.
In simple words: We factored the polynomial into a squared term. Then we found the repeating zero and verified that its sum and product match the coefficient formulas.

🎯 Exam Tip: Recognizing perfect square trinomials can simplify finding zeros. Always remember that a quadratic equation can have two distinct real roots, one repeated real root, or two complex roots.

Question 1. (iii) \( 6x^2 - x - 2 \)
Answer:
For the polynomial \( f(x) = 6x^2 - x - 2 \):
First, factor the polynomial by splitting the middle term:
\( 6x^2 - 4x + 3x - 2 \)
Group terms: \( 2x(3x - 2) + 1(3x - 2) \)
Factor out the common binomial: \( (3x - 2)(2x + 1) \)
To find the zeros, set \( f(x) = 0 \):
\( (3x - 2)(2x + 1) = 0 \)
This means either \( 3x - 2 = 0 \) or \( 2x + 1 = 0 \).
Solving these, we get \( 3x = 2 \implies x = \frac{2}{3} \) or \( 2x = -1 \implies x = -\frac{1}{2} \).
So, the zeros are \( \frac{2}{3} \) and \( -\frac{1}{2} \). Factorizing helps break down a complex polynomial into simpler multiplication.

Now, let's check the relation between the zeros and coefficients:
Sum of zeros \( = \frac{2}{3} + (-\frac{1}{2}) = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \).
Using the formula \( -\frac{b}{a} \): for \( 6x^2 - x - 2 \), \( a=6, b=-1, c=-2 \).
\( -\frac{b}{a} = -\frac{-1}{6} = \frac{1}{6} \). (The sum matches)
Product of zeros \( = \frac{2}{3} \times (-\frac{1}{2}) = -\frac{2}{6} = -\frac{1}{3} \).
Using the formula \( \frac{c}{a} \):
\( \frac{c}{a} = \frac{-2}{6} = -\frac{1}{3} \). (The product matches)
This confirms the relationship between the zeros and coefficients is correct.
In simple words: We broke the polynomial into two parts that multiply together, then found the \( x \) values that make each part zero. We then checked these zeros against the sum and product formulas.

🎯 Exam Tip: When factoring quadratic expressions, look for two numbers that multiply to \( ac \) and add up to \( b \) to split the middle term.

Question 1. (iv) \( x^2 - 15 \)
Answer:
For the polynomial \( f(x) = x^2 - 15 \):
To find the zeros, set \( f(x) = 0 \):
\( x^2 - 15 = 0 \)
\( x^2 = 15 \)
Take the square root of both sides:
\( x = \pm\sqrt{15} \)
So, the zeros are \( \sqrt{15} \) and \( -\sqrt{15} \). This type of factorization is called the difference of squares.

Now, let's check the relation between the zeros and coefficients:
Sum of zeros \( = \sqrt{15} + (-\sqrt{15}) = 0 \).
Using the formula \( -\frac{b}{a} \): for \( x^2 - 15 \), \( a=1, b=0, c=-15 \).
\( -\frac{b}{a} = -\frac{0}{1} = 0 \). (The sum matches)
Product of zeros \( = \sqrt{15} \times (-\sqrt{15}) = -15 \).
Using the formula \( \frac{c}{a} \):
\( \frac{c}{a} = \frac{-15}{1} = -15 \). (The product matches)
This confirms the relationship between the zeros and coefficients is correct.
In simple words: We found the values for \( x \) by isolating \( x^2 \) and taking the square root. We then confirmed that the sum and product of these root values correctly matched what the polynomial's numbers told us.

🎯 Exam Tip: Remember that \( \sqrt{a} \times \sqrt{a} = a \) and \( (\sqrt{a})^2 = a \). Also, don't forget both the positive and negative roots when solving \( x^2 = \text{constant} \).

Question 1. (v) \( x^2 - (\sqrt{3} + 1) x + \sqrt{3} \)
Answer:
For the polynomial \( f(x) = x^2 - (\sqrt{3} + 1) x + \sqrt{3} \):
First, expand the middle term: \( f(x) = x^2 - \sqrt{3}x - x + \sqrt{3} \).
Group terms: \( x(x - \sqrt{3}) - 1(x - \sqrt{3}) \).
Factor out the common binomial: \( (x - \sqrt{3})(x - 1) \).
To find the zeros, set \( f(x) = 0 \):
\( (x - \sqrt{3})(x - 1) = 0 \)
This means either \( x - \sqrt{3} = 0 \) or \( x - 1 = 0 \).
Solving these, we get \( x = \sqrt{3} \) or \( x = 1 \).
So, the zeros are \( \sqrt{3} \) and \( 1 \). This method of factoring helps simplify expressions with square roots.

Now, let's check the relation between the zeros and coefficients:
Sum of zeros \( = \sqrt{3} + 1 \).
Using the formula \( -\frac{b}{a} \): for \( x^2 - (\sqrt{3} + 1) x + \sqrt{3} \), \( a=1, b=-(\sqrt{3} + 1), c=\sqrt{3} \).
\( -\frac{b}{a} = -\frac{-(\sqrt{3} + 1)}{1} = \sqrt{3} + 1 \). (The sum matches)
Product of zeros \( = \sqrt{3} \times 1 = \sqrt{3} \).
Using the formula \( \frac{c}{a} \):
\( \frac{c}{a} = \frac{\sqrt{3}}{1} = \sqrt{3} \). (The product matches)
This confirms the relationship between the zeros and coefficients is correct.
In simple words: We factored the polynomial by grouping terms, found the values of \( x \) that make it zero, and then showed that these zeros correctly fit the formulas for sum and product based on the polynomial's original numbers.

🎯 Exam Tip: When dealing with terms involving square roots, treat them as single entities until you need to combine or factor them. Always be careful with the signs when expanding or factoring.

Question 1. (vi) \( 3x^2 - x - 4 \)
Answer:
For the polynomial \( f(x) = 3x^2 - x - 4 \):
First, factor the polynomial by splitting the middle term:
\( 3x^2 - 4x + 3x - 4 \)
Group terms: \( x(3x - 4) + 1(3x - 4) \)
Factor out the common binomial: \( (3x - 4)(x + 1) \)
To find the zeros, set \( f(x) = 0 \):
\( (3x - 4)(x + 1) = 0 \)
This means either \( 3x - 4 = 0 \) or \( x + 1 = 0 \).
Solving these, we get \( 3x = 4 \implies x = \frac{4}{3} \) or \( x = -1 \).
So, the zeros are \( \frac{4}{3} \) and \( -1 \). Factoring helps simplify the polynomial into easier parts.

Now, let's check the relation between the zeros and coefficients:
Sum of zeros \( = \frac{4}{3} + (-1) = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \).
Using the formula \( -\frac{b}{a} \): for \( 3x^2 - x - 4 \), \( a=3, b=-1, c=-4 \).
\( -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3} \). (The sum matches)
Product of zeros \( = \frac{4}{3} \times (-1) = -\frac{4}{3} \).
Using the formula \( \frac{c}{a} \):
\( \frac{c}{a} = \frac{-4}{3} \). (The product matches)
This confirms the relationship between the zeros and coefficients is correct.
In simple words: We found the values of \( x \) that make the polynomial zero by splitting the middle term and factoring. Then we confirmed that the sum and product of these zeros match the standard formulas.

🎯 Exam Tip: Always double-check your factorization by multiplying the factors back to ensure you get the original polynomial.

Question 2. Find quadratic polynomial. Sum and product of whose zeros are given numbers respectively.
(i) -3, 2
(ii) \( \sqrt{2}, \frac{1}{3} \)
(iii) \( \frac{-1}{4}, \frac{1}{4} \)
(iv) 0, \( \sqrt{5} \)
(v) 4, 1
(vi) 1, 1
Answer:
We use the general formula for a quadratic polynomial when its sum and product of zeros are known: \( f(x) = k\{x^2 - (\text{Sum of Zeros})x + (\text{Product of Zeros})\}\), where \( k \) is any non-zero real number. For the simplest polynomial, we can take \( k=1 \).

(i) Given Sum of Zeros \( = -3 \) and Product of Zeros \( = 2 \).
Substituting these values into the formula:
\( f(x) = k\{x^2 - (-3)x + 2\} \)
\( f(x) = k\{x^2 + 3x + 2\} \)
The simplest quadratic polynomial is \( x^2 + 3x + 2 \). This formula helps us construct an equation from its solutions.
In simple words: We use a special rule that helps us build a quadratic equation when we know the sum and product of its answers (zeros).

🎯 Exam Tip: Always remember the negative sign before the sum of zeros in the formula \( x^2 - (\text{Sum})x + (\text{Product}) \).

Question 2. (ii) \( \sqrt{2}, \frac{1}{3} \)
Answer:
Given Sum of Zeros \( = \sqrt{2} \) and Product of Zeros \( = \frac{1}{3} \).
Substituting these values into the formula:
\( f(x) = k\{x^2 - (\sqrt{2})x + \frac{1}{3}\} \)
To remove the fraction, we can multiply the expression inside the bracket by 3 (or take \( k=3 \)):
\( f(x) = k \left\{ \frac{3x^2 - 3\sqrt{2}x + 1}{3} \right\} \)
\( f(x) = \frac{k}{3} \{3x^2 - 3\sqrt{2}x + 1\} \)
If we choose \( \frac{k}{3} = K \) (another constant), the simplest quadratic polynomial is \( 3x^2 - 3\sqrt{2}x + 1 \). This method cleans up the polynomial by removing fractions.
In simple words: We put the given sum and product of answers into the formula. To make the polynomial look nicer without fractions, we multiplied everything by 3.

🎯 Exam Tip: When you get a polynomial with fractions, you can multiply the entire expression by the least common multiple of the denominators to get integer coefficients, if allowed by the question.

Question 2. (iii) \( \frac{-1}{4}, \frac{1}{4} \)
Answer:
Given Sum of Zeros \( = -\frac{1}{4} \) and Product of Zeros \( = \frac{1}{4} \).
Substituting these values into the formula:
\( f(x) = k\{x^2 - (-\frac{1}{4})x + \frac{1}{4}\} \)
\( f(x) = k\{x^2 + \frac{1}{4}x + \frac{1}{4}\} \)
To remove the fractions, we can multiply the expression inside the bracket by 4 (or take \( k=4 \)):
\( f(x) = k \left\{ \frac{4x^2 + x + 1}{4} \right\} \)
\( f(x) = \frac{k}{4} \{4x^2 + x + 1\} \)
If we choose \( \frac{k}{4} = K \), the simplest quadratic polynomial is \( 4x^2 + x + 1 \). This process simplifies the polynomial's appearance.
In simple words: We used the formula with the given sum and product. Because there were fractions, we multiplied everything by 4 to get whole numbers in the polynomial.

🎯 Exam Tip: Always simplify the polynomial to its integer coefficient form unless specified otherwise, as it makes it easier to work with.

Question 2. (iv) 0, \( \sqrt{5} \)
Answer:
Given Sum of Zeros \( = 0 \) and Product of Zeros \( = \sqrt{5} \).
Substituting these values into the formula:
\( f(x) = k\{x^2 - (0)x + \sqrt{5}\} \)
\( f(x) = k\{x^2 + \sqrt{5}\} \)
The simplest quadratic polynomial is \( x^2 + \sqrt{5} \). When the sum of zeros is zero, it means the roots are opposites, like \( +\sqrt{5} \) and \( -\sqrt{5} \).
In simple words: We used the formula with the given sum and product. Since the sum was zero, the \( x \) term disappeared, leaving a simple polynomial.

🎯 Exam Tip: If the sum of zeros is zero, the coefficient of \( x \) in the polynomial will be zero. If the product of zeros is zero, the constant term will be zero.

Question 2. (v) 4, 1
Answer:
Given Sum of Zeros \( = 4 \) and Product of Zeros \( = 1 \).
Substituting these values into the formula:
\( f(x) = k\{x^2 - (4)x + 1\} \)
\( f(x) = k\{x^2 - 4x + 1\} \)
The simplest quadratic polynomial is \( x^2 - 4x + 1 \). This is a straightforward application of the formula.
In simple words: We directly put the sum and product of the answers into the formula to get the polynomial.

🎯 Exam Tip: Be careful with signs. If the sum of roots is positive, the coefficient of \( x \) in the polynomial will be negative.

Question 2. (vi) 1, 1
Answer:
Given Sum of Zeros \( = 1 \) and Product of Zeros \( = 1 \).
Substituting these values into the formula:
\( f(x) = k\{x^2 - (1)x + 1\} \)
\( f(x) = k\{x^2 - x + 1\} \)
The simplest quadratic polynomial is \( x^2 - x + 1 \). This polynomial has two identical roots, meaning the graph would just touch the x-axis at one point.
In simple words: We used the formula with the sum and product both being 1 to find the polynomial.

🎯 Exam Tip: When the sum and product of zeros are simple integers like 1, the calculation is often less prone to error, but sign awareness is still key.

Question 3. If sum of square of zeros of quadratic equation \( f(x) = x^2 - 8x + k \) is 40 then find the value of k.
Answer:
Given the quadratic polynomial \( f(x) = x^2 - 8x + k \).
Let the zeros of the polynomial be \( \alpha \) and \( \beta \).
From the relations between zeros and coefficients for \( ax^2 + bx + c \):
Sum of zeros: \( \alpha + \beta = -\frac{b}{a} \)
Here, \( a=1, b=-8, c=k \).
So, \( \alpha + \beta = -\frac{(-8)}{1} = 8 \). (Equation 1)

Product of zeros: \( \alpha \beta = \frac{c}{a} \)
So, \( \alpha \beta = \frac{k}{1} = k \). (Equation 2)

We are given that the sum of the square of zeros is 40:
\( \alpha^2 + \beta^2 = 40 \).

We know the identity: \( (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \). This identity is very useful for relating sums and products of roots to their squares.

Substitute the known values into the identity:
From Equation 1, \( \alpha + \beta = 8 \), so \( (\alpha + \beta)^2 = 8^2 = 64 \).
From the problem, \( \alpha^2 + \beta^2 = 40 \).
From Equation 2, \( \alpha \beta = k \).

Now, put these into the identity:
\( 64 = 40 + 2k \)
Subtract 40 from both sides:
\( 2k = 64 - 40 \)
\( 2k = 24 \)
Divide by 2:
\( k = 12 \).
Thus, the value of \( k \) is 12.
In simple words: We used the relationships between the zeros (answers) and the numbers in the polynomial. We were given that the sum of the squares of the zeros is 40. By using an algebraic identity, we could set up an equation to find the missing value \( k \).

🎯 Exam Tip: Make sure you know the identities \( (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \) and \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) by heart, as they are key for solving such problems.