CBSE Class 10 Mathematics Polynomials Notes Set 12

Polynomials and Their Degree

A polynomial in x is a sum of terms of the form a·xⁿ with n a whole number. The highest power present is its degree. Degree 1, 2 and 3 polynomials are called linear, quadratic and cubic. A constant has degree 0.
A number k is a zero of p(x) when p(k) = 0. A polynomial of degree n has at most n real zeros.

Remainder Theorem and Factor Theorem

Remainder theorem: when p(x) is divided by (x − k), the remainder equals p(k). For a divisor (ax − b) the remainder is p(b/a).
Factor theorem: (x − k) is a factor of p(x) exactly when p(k) = 0.
If the divisor is quadratic, the remainder has the form ax + b; use the two zeros of the divisor to find a and b.

Division Algorithm

For polynomials p(x) and g(x) with g(x) not zero there are polynomials q(x) and r(x) with p(x) = g(x)·q(x) + r(x), where r(x) = 0 or the degree of r(x) is less than the degree of g(x). Divide term by term, always matching the highest power, and stop when the degree of the remainder drops below that of the divisor.

Algebraic Identities

(a ± b)² = a² ± 2ab + b². a² − b² = (a − b)(a + b).
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³ = a³ ± b³ ± 3ab(a ± b).
a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). In particular, if a + b + c = 0 then a³ + b³ + c³ = 3abc.
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Zeros and Coefficients

For ax² + bx + c with zeros α and β: α + β = −b/a and αβ = c/a.
For ax³ + bx² + cx + d with zeros α, β, γ: α + β + γ = −b/a, αβ + βγ + γα = c/a and αβγ = −d/a.
Symmetric expressions such as α² + β² = (α + β)² − 2αβ and 1/α + 1/β = (α + β)/(αβ) can be found without solving for the zeros.

Graph of a Quadratic Polynomial

The graph of y = ax² + bx + c is a parabola. It opens upwards if a > 0 (the vertex is the lowest point) and downwards if a < 0 (the vertex is the highest point). The vertex lies at x = −b/(2a), and the number of times the parabola meets the x-axis equals the number of real zeros.

Solved Examples

Example 1. Find the remainder when x³ − 6x² + 11x − 2 is divided by (x − 3).
Answer:
Put x = 3 in x³ − 6x² + 11x − 2: 27 − 54 + 33 − 2 = 4.
Remainder = 4

Example 2. For what value of k is (x + 2) a factor of x³ + 3x² − kx + 6?
Answer:
Put x = −2: −8 + 12 + 2k + 6 = 0, so 2k = −10.
k = −5

Example 3. Find the remainder when x¹⁰⁰ − 1 is divided by x + 1.
Answer:
Put x = −1: (−1)¹⁰⁰ − 1 = 1 − 1 = 0.
Remainder = 0, so x + 1 is a factor.

Example 4. If x + 1/x = 4, find x³ + 1/x³.
Answer:
x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 64 − 12.
52

Example 5. Find the value of 25³ + (−12)³ + (−13)³ using an identity.
Answer:
25 − 12 − 13 = 0, so the sum equals 3 × 25 × (−12) × (−13).
3 × 3900 = 11700

Example 6. Factorise x³ − 7x + 6.
Answer:
x = 1 is a zero (1 − 7 + 6 = 0), so divide by (x − 1) to get x² + x − 6 = (x + 3)(x − 2).
(x − 1)(x − 2)(x + 3)

Example 7. The zeros of 2x² − 5x + 1 are α and β. Find α² + β².
Answer:
α + β = 5/2 and αβ = 1/2.
α² + β² = 25/4 − 1 = 21/4.

Example 8. Divide 6x³ + x² − 5x + 2 by 2x + 1 and state the quotient and remainder.
Answer:
6x³ ÷ 2x = 3x²; subtract 3x²(2x + 1) to leave −2x² − 5x + 2. Then −2x² ÷ 2x = −x; subtract to leave −4x + 2. Then −4x ÷ 2x = −2; subtract to leave 4.
Quotient 3x² − x − 2 and remainder 4.

Example 9. Find a cubic polynomial whose zeros are 1, 2 and 3.
Answer:
Sum = 6, sum of products = 11, product = 6.
x³ − 6x² + 11x − 6

Example 10. Find the minimum value of x² − 4x + 7.
Answer:
x² − 4x + 7 = (x − 2)² + 3, and a square is never negative.
Minimum value 3 at x = 2.