GSEB Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.4

Question 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
1. \( 2x^3 + x^2 – 5x + 2; \frac{1}{2}, 1, -2 \)
2. \( x^3 – 4x^2 + 5x – 2; 2, 1, 1 \)
Answer:
1. Let \( p(x) = 2x^3 + x^2 – 5x + 2 \)
Then, we have
\( p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{2}-5\left(\frac{1}{2}\right)+2 \)
\( = \frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0 \)
\( p(1) = 2(1)^3 + (1)^2 – 5(1) + 2 \)
\( = 2+1-5+2 = 0 \)
and,
\( p(-2) = 2(-2)^3 + (-2)^2 – 5(-2) + 2 \)
\( = -16 + 4 + 10 + 2 = 0 \)
So, \( \frac{1}{2}, 1 \) and \( -2 \) are the zeroes of \( 2x^3 + x^2 – 5x + 2 \). This means when you substitute these values into the polynomial, the result is zero.
Comparing the given polynomial with \( ax^3 + bx^2 + cx + d \), we get
\( a = 2 \)
\( b = 1 \)
\( c = -5 \)
\( d = 2 \)
Let \( \alpha = \frac{1}{2} \)
\( \beta = 1 \)
and \( \gamma = -2 \)
Then, we have
\( \alpha + \beta + \gamma = \frac{1}{2} + 1 + (-2) \)
\( = \frac{1}{2} - 1 = -\frac{1}{2} \)
We also know that \( \alpha + \beta + \gamma = -\frac{b}{a} \)
\( = -\frac{1}{2} \)
So, this matches.

\( \alpha\beta + \beta\gamma + \gamma\alpha = \left(\frac{1}{2}\right) (1) + (1)(-2) + (-2)\left(\frac{1}{2}\right) \)
\( = \frac{1}{2} - 2 - 1 \)
\( = \frac{1}{2} - 3 \)
\( = \frac{1-6}{2} = -\frac{5}{2} \)
We also know that \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
\( = \frac{-5}{2} \)
This also matches.

\( \alpha\beta\gamma = \left(\frac{1}{2}\right) (1) (-2) \)
\( = -1 \)
We also know that \( \alpha\beta\gamma = -\frac{d}{a} \)
\( = -\frac{2}{2} = -1 \)
This relationship also holds true.

2. Let \( p(x) = x^3 – 4x^2 + 5x – 2 \)
Then, we have
\( p(2) = (2)^3 – 4(2)^2 + 5(2) – 2 \)
\( = 8 - 16 + 10 - 2 = 0 \)
\( p(1) = (1)^3 – 4(1)^2 + 5(1) – 2 \)
\( = 1 - 4 + 5 - 2 = 0 \)
Therefore, 2 and 1 are the two zeroes of \( x^3 – 4x^2 + 5x – 2 \). This means \( (x – 2) (x – 1) \), which is \( x^2 – 3x + 2 \), is a factor of the given polynomial. Now, we use the division algorithm to divide the polynomial by \( x^2 – 3x + 2 \).
\[ x^2-3x + 2 \overline{) x^3 - 4x^2 + 5x - 2 } \] \[ \quad \quad \quad \underline{x^3 - 3x^2 + 2x} \] \[ \quad \quad \quad \quad \quad -x^2 + 3x - 2 \] \[ \quad \quad \quad \quad \quad \underline{-x^2 + 3x - 2} \] \[ \quad \quad \quad \quad \quad \quad \quad \quad 0 \] So, \( x^3 – 4x^2 + 5x - 2 \)
\( = (x^2 – 3x + 2) (x – 1) \)
\( \implies x^3 – 4x^2 + 5x - 2 = (x – 2) (x – 1) (x – 1) \)
Hence, 2, 1 and 1 are the zeroes of \( x^3 – 4x^2 + 5x - 2 \). These are the values that make the polynomial equal to zero.
Comparing the given polynomial with \( ax^3 + bx^2 + cx + d \), we get
\( a = 1 \)
\( b = -4 \)
\( c = 5 \)
\( d = -2 \)
Let \( \alpha = 2 \)
\( \beta = 1 \)
and \( \gamma = 1 \)
Then, we have
\( \alpha + \beta + \gamma = 2 + 1 + 1 = 4 \)
\( = -\frac{-4}{1} = -\frac{b}{a} \)
The sum of the zeroes matches.

\( \alpha\beta + \beta\gamma + \gamma\alpha = (2)(1) + (1)(1) + (1)(2) \)
\( = 2 + 1 + 2 = 5 \)
\( = \frac{5}{1} = \frac{c}{a} \)
The sum of the products of zeroes taken two at a time also matches.

\( \alpha\beta\gamma = (2) (1) (1) = 2 \)
\( = -\frac{-2}{1} = -\frac{d}{a} \)
The product of the zeroes also matches the formula.
In simple words: We checked if the given numbers made the polynomial zero. Then, we compared the sum, product, and sum of pairwise products of these zeroes with the formulas \( -\frac{b}{a} \), \( \frac{c}{a} \), and \( -\frac{d}{a} \) respectively. All the relationships held true.

Exam Tip: Remember to perform polynomial division carefully, ensuring all terms are aligned and signs are correctly handled during subtraction to avoid errors.

Question 2. Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Answer: Let the cubic polynomial be \( ax^3 + bx^2 + cx + d \) and its zeroes be \( \alpha, \beta \) and \( \gamma \).
We are given:
Sum of zeroes: \( \alpha + \beta + \gamma = 2 \)
We know \( \alpha + \beta + \gamma = -\frac{b}{a} \).
So, \( -\frac{b}{a} = 2 \)
Sum of the product of its zeroes taken two at a time: \( \alpha\beta + \beta\gamma + \gamma\alpha = -7 \)
We know \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \).
So, \( \frac{c}{a} = -7 \)
Product of its zeroes: \( \alpha\beta\gamma = -14 \)
We know \( \alpha\beta\gamma = -\frac{d}{a} \).
So, \( -\frac{d}{a} = -14 \)
Let's take \( a = 1 \) for simplicity. Then we can find the other coefficients:
\( -\frac{b}{1} = 2 \implies b = -2 \)
\( \frac{c}{1} = -7 \implies c = -7 \)
\( -\frac{d}{1} = -14 \implies d = 14 \)
Hence, one cubic polynomial which fits the given conditions is \( x^3 – 2x^2 – 7x + 14 \). This polynomial satisfies all the stated properties for its zeroes.
In simple words: We used the given sum, sum of pairwise products, and product of zeroes to find the coefficients of a cubic polynomial. By setting 'a' to 1, we determined 'b', 'c', and 'd' and then constructed the polynomial.

Exam Tip: Remember the standard formulas for the sum, sum of products, and product of zeroes of a cubic polynomial to quickly find the coefficients.

Question 3. If the zeroes of the polynomial \( x^3 - 3x^2 + x + 1 \) are a – b, a, a + b, find a and b.
Answer: The given polynomial is \( x^3 - 3x^2 + x + 1 \).
Comparing with the standard cubic polynomial \( Ax^3 + Bx^2 + Cx + D \), we get:
\( A = 1 \)
\( B = -3 \)
\( C = 1 \)
\( D = 1 \)
Let the zeroes be \( \alpha = a - b \), \( \beta = a \), and \( \gamma = a + b \).
Then, we use the relationships between the zeroes and coefficients:
Sum of the zeroes: \( \alpha + \beta + \gamma = -\frac{B}{A} \)
\( (a - b) + a + (a + b) = -\frac{(-3)}{1} \)
\( 3a = 3 \)
\( \implies a = 1 \)

Product of the zeroes: \( \alpha\beta\gamma = -\frac{D}{A} \)
\( (a – b) a (a + b) = -\frac{1}{1} \)
\( a(a^2 – b^2) = -1 \)
Substitute the value \( a = 1 \):
\( 1(1^2 – b^2) = -1 \)
\( 1 – b^2 = -1 \)
\( -b^2 = -1 - 1 \)
\( -b^2 = -2 \)
\( b^2 = 2 \)
\( \implies b = \pm \sqrt {2} \)
So, the values are \( a = 1 \) and \( b = \pm \sqrt {2} \). These are the numbers we need to find.
In simple words: We used the properties that connect the zeroes of a polynomial to its coefficients. By adding the zeroes together and multiplying them, we created equations that helped us solve for the unknown values 'a' and 'b'.

Exam Tip: For problems involving zeroes in arithmetic progression, using the sum of roots is often the easiest way to find 'a'. Then use the product of roots to find 'b'.

Question 4. If two zeroes of the polynomial \( x^4 – 6x^3 – 26x^2 + 138x – 35 \) are \( 2 \pm \sqrt {3} \), find the other zeroes.
Answer: Since two zeroes of the given polynomial \( x^4 – 6x^3 – 26x^2 + 138x – 35 \) are \( 2 + \sqrt {3} \) and \( 2 - \sqrt {3} \),
Therefore, a factor of the polynomial is:
\( (x – (2 + \sqrt {3} )) (x – (2 – \sqrt {3} )) \)
\( = ((x - 2) - \sqrt {3}) ((x - 2) + \sqrt {3}) \)
This is in the form \( (A - B)(A + B) = A^2 - B^2 \), where \( A = (x - 2) \) and \( B = \sqrt {3} \).
So, \( (x – 2)^2 - (\sqrt {3})^2 \)
\( = (x^2 – 4x + 4) – 3 \)
\( = x^2 – 4x + 1 \)
Thus, \( x^2 – 4x + 1 \) is a factor of the given polynomial. Now, we apply the division algorithm to divide the polynomial by this factor.
\[ x^2-4x+1 \overline{) x^4 - 6x^3 - 26x^2 + 138x - 35 } \] \[ \quad \quad \quad \underline{x^4 - 4x^3 + x^2} \] \[ \quad \quad \quad \quad \quad -2x^3 - 27x^2 + 138x - 35 \] \[ \quad \quad \quad \quad \quad \underline{-2x^3 + 8x^2 - 2x} \] \[ \quad \quad \quad \quad \quad \quad \quad -35x^2 + 140x - 35 \] \[ \quad \quad \quad \quad \quad \quad \quad \underline{-35x^2 + 140x - 35} \] \[ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0 \] So, \( x^4 – 6x^3 – 26x^2 + 138x – 35 \)
\( = (x^2 – 4x + 1) (x^2 – 2x – 35) \)
Now, we need to find the zeroes of the quadratic factor \( x^2 – 2x – 35 \).
\( x^2 – 2x – 35 = 0 \)
We can factor this quadratic: find two numbers that multiply to -35 and add to -2. These are -7 and 5.
\( (x – 7) (x + 5) = 0 \)
This gives us \( x – 7 = 0 \implies x = 7 \)
and \( x + 5 = 0 \implies x = -5 \)
Hence, the other two zeroes are 7 and -5. These are the remaining roots of the polynomial.
In simple words: Since we knew two zeroes, we formed a quadratic factor from them. We then divided the main polynomial by this factor. The quotient was another quadratic equation, which we solved to find the two missing zeroes.

Exam Tip: When given irrational or conjugate zeroes, always form a quadratic factor first using \( (x - \text{zero}_1)(x - \text{zero}_2) \) before performing polynomial division. This is a common strategy for finding remaining roots.

Question 5. If the polynomial \( x^4 – 6x^3 + 16x^2 – 25x + 10 \) is divided by another polynomial \( x^2 – 2x + k \), the remainder comes out to be \( x + a \), find k and a.
Answer: Let us apply the division algorithm to the given Polynomial \( x^4 – 6x^3 + 16x^2 – 25x + 10 \) and another Polynomial \( x^2 – 2x + k \).
\[ x^2-2x+k \overline{) x^4 - 6x^3 + 16x^2 - 25x + 10 } \] \[ \quad \quad \quad \underline{x^4 - 2x^3 + kx^2} \] \[ \quad \quad \quad \quad \quad -4x^3 + (16-k)x^2 - 25x + 10 \] \[ \quad \quad \quad \quad \quad \underline{-4x^3 + 8x^2 - 4kx} \] \[ \quad \quad \quad \quad \quad \quad \quad (16-k-8)x^2 + (-25+4k)x + 10 \] \[ \quad \quad \quad \quad \quad \quad \quad (8-k)x^2 + (4k-25)x + 10 \] \[ \quad \quad \quad \quad \quad \quad \quad \underline{(8-k)x^2 - 2(8-k)x + k(8-k)} \] \[ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (-25+4k+16-2k)x + (10-k(8-k)) \] \[ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2k-9)x + (10-8k+k^2) \] The remainder is \( (2k – 9)x + (k^2 – 8k + 10) \).
But the remainder is given to be \( x + a \).
We compare the coefficients of \( x \) and the constant terms in both remainders.
Comparing the coefficient of \( x \):
\( 2k – 9 = 1 \)
\( 2k = 1 + 9 \)
\( 2k = 10 \)
\( k = 5 \)

Comparing the constant terms:
\( k^2 – 8k + 10 = a \)
Substitute the value of \( k = 5 \) into this equation:
\( (5)^2 – 8(5) + 10 = a \)
\( 25 – 40 + 10 = a \)
\( -15 + 10 = a \)
\( -5 = a \)
\( \implies a = -5 \)
So, the values are \( k = 5 \) and \( a = -5 \). These are the constants we needed to find.
In simple words: We performed polynomial division to find the remainder. Then, we matched the terms of our calculated remainder with the given remainder \( x + a \) to set up equations. Solving these equations helped us determine the values of 'k' and 'a'.

Exam Tip: Be very careful with signs and combining like terms during polynomial long division, especially when coefficients involve variables like 'k'. Double-check each step to ensure accuracy.