CBSE Class 10 Mathematics Polynomials Notes Set 08

Relationship between Zeros and Coefficients of a Quadratic Polynomial

Let \( \alpha \) and \( \beta \) be the zeros of a quadratic polynomial \( f(x) = ax^2 + bx + c \), \( a \ne 0 \). By the factor theorem, \( (x - \alpha) \) and \( (x - \beta) \) are its factors, so \( f(x) = k(x - \alpha)(x - \beta) \) for some non-zero constant k.

\( ax^2 + bx + c = k\{x^2 - (\alpha + \beta)x + \alpha\beta\} = kx^2 - k(\alpha + \beta)x + k\alpha\beta \).
Comparing the coefficients of \( x^2 \), x and the constant terms: \( a = k \), \( b = -k(\alpha + \beta) \) and \( c = k\alpha\beta \). Hence

\[ \alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}, \qquad \alpha\beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}. \]

Remark. If \( \alpha \) and \( \beta \) are the zeros of a quadratic polynomial f(x), then \( f(x) = k\{x^2 - (\alpha + \beta)x + \alpha\beta\} \), or \( f(x) = k\{x^2 - (\text{sum of the zeros})x + \text{product of the zeros}\} \), where k is any non-zero real number.

Solved Examples on Zeros and Coefficients

Example 1. Find the zeros of the quadratic polynomial \( f(x) = x^2 - 2x - 8 \) and verify the relationship between the zeros and their coefficients.
Answer:
\( f(x) = x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2) \). Zeros: \( f(x) = 0 \) gives x = 4 or x = −2, so \( \alpha = 4 \) and \( \beta = -2 \).
Sum of zeros = 4 + (−2) = 2, and \( -\dfrac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\dfrac{-2}{1} = 2 \). Verified.
Product of zeros = (4)(−2) = −8, and \( \dfrac{\text{Constant term}}{\text{Coefficient of } x^2} = \dfrac{-8}{1} = -8 \). Verified.

Example 2. Find a quadratic polynomial whose zeros are \( 5 + \sqrt{2} \) and \( 5 - \sqrt{2} \).
Answer:
Sum of zeros = \( 5 + \sqrt{2} + 5 - \sqrt{2} = 10 \). Product = \( (5 + \sqrt{2})(5 - \sqrt{2}) = 25 - 2 = 23 \).
\( f(x) = k\{x^2 - 10x + 23\} \), where k is any non-zero real number.

Example 3. The sum and product of the zeros of a quadratic polynomial are 5 and 17 respectively. Find the polynomial.
Answer:
\( f(x) = k\{x^2 - (\text{sum of zeros})x + \text{product of zeros}\} = k\{x^2 - 5x + 17\} \), where k is any non-zero real number.

Example 4. Form the quadratic polynomial whose zeros are the squares of the zeros of the polynomial \( x^2 - 2x - 15 \).
Answer:
\( x^2 - 2x - 15 = x^2 - 5x + 3x - 15 = (x - 5)(x + 3) \), so the zeros are 5 and −3.
Squares of the zeros: \( 5^2 = 25 \) and \( (-3)^2 = 9 \). Sum = 34 and product = 225.
The required polynomial is \( x^2 - 34x + 225 \).

Check Your Progress: Zeros and Coefficients

Check Your Progress 1. Find the polynomial whose zeros are the squares of the zeros of the polynomial \( x^2 + x - 30 \).
Answer:
\( x^2 + x - 30 = (x + 6)(x - 5) \), zeros 5 and −6. Squares: 25 and 36 with sum 61 and product 900. Polynomial: \( x^2 - 61x + 900 \).

Check Your Progress 2. Find the polynomial whose zeros are double the zeros of the polynomial \( x^2 - x - 42 \).
Answer:
\( x^2 - x - 42 = (x - 7)(x + 6) \), zeros 7 and −6. Doubles: 14 and −12 with sum 2 and product −168. Polynomial: \( x^2 - 2x - 168 \).

Check Your Progress 3. Find the polynomial whose zeros are the reciprocals of the zeros of the polynomial \( 2x^2 + 13x + 15 \).
Answer:
\( 2x^2 + 13x + 15 = (2x + 3)(x + 5) \), zeros \( -\tfrac{3}{2} \) and −5. Reciprocals: \( -\tfrac{2}{3} \) and \( -\tfrac{1}{5} \) with sum \( -\tfrac{13}{15} \) and product \( \tfrac{2}{15} \). Polynomial: \( 15x^2 + 13x + 2 \).

Check Your Progress 4. Find the polynomial whose zeros are one more than the zeros of the polynomial \( 6x^2 + 17x + 12 \).
Answer:
\( 6x^2 + 17x + 12 = (3x + 4)(2x + 3) \), zeros \( -\tfrac{4}{3} \) and \( -\tfrac{3}{2} \). One more: \( -\tfrac{1}{3} \) and \( -\tfrac{1}{2} \) with sum \( -\tfrac{5}{6} \) and product \( \tfrac{1}{6} \). Polynomial: \( 6x^2 + 5x + 1 \).

Division Algorithm for Polynomials

If p(x) and g(x) are any two polynomials with \( g(x) \ne 0 \), then we can find polynomials q(x) and r(x) such that \( p(x) = g(x) \times q(x) + r(x) \), that is, Dividend = Divisor × Quotient + Remainder, where \( r(x) = 0 \) or the degree of r(x) is less than the degree of g(x).

1. If r(x) = 0, then g(x) is a factor of p(x).
2. If the degree of p(x) is greater than the degree of g(x), then deg q(x) = deg p(x) − deg g(x).
3. If the degree of p(x) equals the degree of g(x), then deg q(x) = 0 and deg r(x) is less than deg g(x).

Solved Examples on the Division Algorithm

Example 5. What must be added to \( 3x^3 + x^2 - 22x + 9 \) so that the result is exactly divisible by \( 3x^2 + 7x - 6 \)?
Answer:
Let the polynomial to be added be \( ax + b \) (the remainder has degree less than 2). Then \( s(x) = 3x^3 + x^2 - (22 - a)x + (9 + b) \) must be exactly divisible by \( 3x^2 + 7x - 6 \).
Long division: the first quotient term is x, since \( x(3x^2 + 7x - 6) = 3x^3 + 7x^2 - 6x \); the remainder so far is \( -6x^2 + (a - 16)x + 9 + b \). The next quotient term is −2, since \( -2(3x^2 + 7x - 6) = -6x^2 - 14x + 12 \). The final remainder is \( (a - 2)x + (b - 3) \).
For exact divisibility: \( a - 2 = 0 \) and \( b - 3 = 0 \), so a = 2 and b = 3. The quotient is x − 2.
2x + 3 must be added.

Example 6. What must be subtracted from \( x^3 - 6x^2 - 15x + 80 \) so that the result is exactly divisible by \( x^2 + x - 12 \)?
Answer:
Let \( ax + b \) be subtracted. \( s(x) = x^3 - 6x^2 - (15 + a)x + (80 - b) \) must be divisible by \( x^2 + x - 12 \).
Quotient x − 7, since \( (x - 7)(x^2 + x - 12) = x^3 - 6x^2 - 19x + 84 \). Comparing: \( 15 + a = 19 \) and \( 80 - b = 84 \), so a = 4 and b = −4.
4x − 4 must be subtracted.

Example 7. Apply the division algorithm to find the quotient q(x) and remainder r(x) on dividing \( f(x) = 10x^4 + 17x^3 - 62x^2 + 30x - 3 \) by \( g(x) = 2x^2 - x + 1 \).
Answer:
Step 1: \( 5x^2 \times (2x^2 - x + 1) = 10x^4 - 5x^3 + 5x^2 \); subtract to get \( 22x^3 - 67x^2 + 30x - 3 \).
Step 2: \( 11x \times (2x^2 - x + 1) = 22x^3 - 11x^2 + 11x \); subtract to get \( -56x^2 + 19x - 3 \).
Step 3: \( -28 \times (2x^2 - x + 1) = -56x^2 + 28x - 28 \); subtract to get \( -9x + 25 \).
So the quotient is \( q(x) = 5x^2 + 11x - 28 \) and the remainder is \( r(x) = -9x + 25 \).
Check: \( (5x^2 + 11x - 28)(2x^2 - x + 1) + (-9x + 25) = 10x^4 + 17x^3 - 62x^2 + 30x - 3 \). The division algorithm is verified.

Example 8. If 1 and −2 are zeros of \( x^4 - 4x^3 - x^2 + 16x - 12 \), find the other zeros.
Answer:
The quadratic polynomial with zeros 1 and −2 is \( (x - 1)(x + 2) = x^2 + x - 2 \). Dividing \( x^4 - 4x^3 - x^2 + 16x - 12 \) by \( x^2 + x - 2 \) gives the quotient \( x^2 - 5x + 6 \) with remainder 0.
\( x^2 - 5x + 6 = (x - 2)(x - 3) \), so the other two zeros are 2 and 3.

Example 9. Find all the zeros of \( f(x) = 2x^4 - 2x^3 - 7x^2 + 3x + 6 \), if two of its zeros are \( -\sqrt{\dfrac{3}{2}} \) and \( \sqrt{\dfrac{3}{2}} \).
Answer:
\( \left(x + \sqrt{\dfrac{3}{2}}\right)\left(x - \sqrt{\dfrac{3}{2}}\right) = x^2 - \dfrac{3}{2} = \dfrac{2x^2 - 3}{2} \), so \( 2x^2 - 3 \) is a factor of f(x).
Dividing, \( 2x^4 - 2x^3 - 7x^2 + 3x + 6 = (2x^2 - 3)(x^2 - x - 2) \) with remainder 0.
\( x^2 - x - 2 = (x - 2)(x + 1) \). So the zeros are \( -\sqrt{\tfrac{3}{2}}, \sqrt{\tfrac{3}{2}}, 2 \) and \( -1 \).