CBSE Class 10 Mathematics Polynomials Notes Set 06

Polynomials

An algebraic expression p(x) of the form \( p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \), where \( a_0, a_1, \ldots, a_n \) are real numbers and all the powers of x are non-negative integers, is called a polynomial. The word polynomial comes from the Greek words poly (many) and nomial (terms), so polynomial means many terms.

Examples: (i) \( x^2 + 9x \) (ii) \( 4x^5 - \sqrt{5}x^2 + 8 \) (iii) \( 3x^3 + \frac{5}{8}x + 5 \).
Not polynomials: (i) \( 3x - \frac{2}{x} - 5 \), because the power of x is −1 (negative); (ii) \( 4x^3 - 3\sqrt{x} + 6 \), because the power of x is \( \frac{1}{2} \) (a fraction).

Degree of a Polynomial

The highest power (exponent) of the terms of a polynomial is called the degree of the polynomial. In \( 4x^3 - 7x^6 + 4 \) the power of \( 4x^3 \) is 3, of \( -7x^6 \) is 6 and of 4 is 0, so the degree is 6. More examples: the degree of \( 5y^2 - 6 \) is 2; of \( 7x - 13x^5 + 4x^3 \) is 5; of 13x + 7 is 1, as x = x¹.

Mind it
1. The degree of a polynomial in more than one variable is found by adding the exponents of the variables in each term. For \( 7a^4 - 3ab^4 \): the term \( 7a^4 \) has degree 4 and the term \( -3ab^4 \) has degree 1 + 4 = 5.

Types of Polynomials

Polynomials are classified in three ways: by degree, by number of terms and by number of distinct variables.

Based on degree

A polynomial of degree n, for n greater than 3, is called a polynomial of degree n, although quartic and quintic are sometimes used for degree 4 and 5.

Based on number of terms

Based on number of distinct variables

In this chapter we mainly discuss polynomials in one variable. Examples: \( P(x) = 4x^7 - 3x^2 + 5 \) is a trinomial of degree 7; \( Q(y) = 5y^2 + 9 \) is a quadratic binomial; R(s) = 75 is a constant (degree 0) monomial. In general, a polynomial in one variable x of degree n is \( a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \), with \( a_n \neq 0 \), where \( a_n, \ldots, a_0 \) are constants.

Values and Zeroes of a Polynomial

If p(x) is a polynomial in x and a is any real number, the value obtained by replacing x by a in p(x) is called the value of p(x) at x = a and is denoted by p(a). For example, for \( f(x) = 2x^3 - 5x^2 + 7 \), f(5) = 2(5)³ − 5(5)² + 7 = 250 − 125 + 7 = 132.

If p(a) = 0, then x = a is a zero of the polynomial p(x). For p(x) = x − 2, p(2) = 0, so x = 2 (or simply 2) is a zero of p(x).

1. A polynomial of degree n can have at most n zeroes. A polynomial of degree 5 has at most 5 zeroes.
2. Difference between zeroes and solutions. Draw the graph of \( f(x) = x^2 - 2x - 8 \) using the table of values below. All the points on the curve are solutions of y = f(x), but only the points where y = 0, that is (−2, 0) and (4, 0), give the zeroes of f(x): −2 and 4.

Factor Theorem

Let p(x) be a polynomial of degree n ≥ 1 and a be a real number such that p(a) = 0. Then (x − a) is a factor of p(x). Conversely, if (x − a) is a factor of p(x), then p(a) = 0.

Remainder Theorem

Let p(x) be any polynomial of degree n ≥ 1 and a be any real number. If p(x) is divided by (x − a), then the remainder is equal to p(a). (If p(a) = 0, the remainder is zero and (x − a) is a factor.)

Geometrical Meaning of the Zeroes of a Polynomial

When the graph of a polynomial is drawn on graph paper, it cuts or touches the x-axis at as many points as there are zeroes of the polynomial. The zeroes of f(x) are the x-coordinates of the points where the graph of y = f(x) intersects the x-axis.

Graph of a Linear Polynomial

The graph of a linear polynomial f(x) = ax + b, a ≠ 0, is always a straight line. Two points are enough to draw it. The line y = ax + b crosses the x-axis at exactly one point, namely \( \left(-\frac{b}{a}, 0\right) \).

Graph of a Quadratic Polynomial

The graph of any quadratic polynomial \( f(x) = ax^2 + bx + c \), a ≠ 0, is a parabola, a curved shape opening upwards if a > 0 and downwards if a < 0.

Let D = b² − 4ac (the discriminant).

Case I: If D > 0, the parabola intersects the x-axis at two distinct points, so the quadratic polynomial \( ax^2 + bx + c \) has 2 zeroes. The zeroes are real and distinct.
Case II: If D = 0, the parabola just touches the x-axis at one point, so the polynomial has 1 zero. The zeroes are real and equal.
Case III: If D < 0, the parabola does not intersect the x-axis at all, so the polynomial has no real zero (the roots are imaginary).

Here α and β are the zeroes of the quadratic polynomial. You will study the discriminant in detail in the chapter on Quadratic Equations.

Graph of a Cubic Polynomial

Graphs of cubic polynomials do not have a fixed shape. The graph of a cubic polynomial always crosses the x-axis at least once and at most thrice.

Mind it
1. A non-zero constant is a polynomial of degree zero, but the degree of the zero polynomial is not defined.
2. A zero of a polynomial is the x-coordinate of a point where the curve y = f(x) meets the line y = 0 (the x-axis).
3. A polynomial of degree n ≥ 1 can have at most n real zeroes.
4. A non-zero constant polynomial has no zeroes.
5. If the sum of the coefficients of a polynomial is zero, then (x − 1) is a factor of the polynomial.

Examples on Degree, Zeroes and the Remainder Theorem

Example 1. The degree of the polynomial 5 is ________?
Answer:
The degree is the highest power of the variable. Write 5 = 5 × x⁰ (since x⁰ = 1). The highest power of x is 0.
Hence the degree of the polynomial 5 is 0.

Example 2. Find the remainder if \( f(x) = x^4 - 3x^2 + 4 \) is divided by g(x) = x − 2.
Answer:
By the remainder theorem the remainder is f(2) = 2⁴ − 3(2)² + 4 = 16 − 12 + 4 = 8.

Example 3. Find the remainder if \( f(x) = 4x^3 - 12x^2 + 14x - 3 \) is divided by g(x) = 2x − 1.
Answer:
g(x) = 2\( \left(x - \frac{1}{2}\right) \), so the remainder is \( f\left(\frac{1}{2}\right) \).
\( f\left(\frac12\right) = 4 \cdot \frac18 - 12 \cdot \frac14 + 14 \cdot \frac12 - 3 = \frac12 - 3 + 7 - 3 = \frac32 \).
Remainder = \( \frac{3}{2} \).

Example 4. The graphical representation of the quadratic polynomial y = ax² + bx + c is shown in the given figure. Does it have (i) any solution? (ii) zero(s)?
Answer:

(i) There are infinitely many solutions, because infinitely many points satisfy the equation y = ax² + bx + c.
(ii) There is no zero, because graphically there is no point of intersection of the x-axis and the curve.

Example 5. Find the remainder when \( x^3 - 6x^2 + 2x - 4 \) is divided by 3x − 1.
Answer:
Let \( f(x) = x^3 - 6x^2 + 2x - 4 \). Since 3x − 1 = 3\( \left(x - \frac13\right) \), the remainder is \( f\left(\frac13\right) \).
\( f\left(\frac13\right) = \frac{1}{27} - \frac{6}{9} + \frac{2}{3} - 4 = \frac{1 - 18 + 18 - 108}{27} = -\frac{107}{27} \).
Remainder = \( -\frac{107}{27} \).

Example 6. Find the values of a and b if \( x^3 - ax^2 - 13x + b \) has (x − 1) and (x + 3) as factors.
Answer:
Let f(x) = x³ − ax² − 13x + b. By the factor theorem f(1) = 0 and f(−3) = 0.
f(1) = 1 − a − 13 + b = 0, so a − b = −12. ... (i)
f(−3) = −27 − 9a + 39 + b = 0, so 9a − b = 12. ... (ii)
Subtracting (i) from (ii): 8a = 24, so a = 3, and then b = a + 12 = 15.
a = 3, b = 15

Example 7. The polynomials \( az^3 + 4z^2 + 3z - 4 \) and \( z^3 - 4z + a \) leave the same remainder when divided by z − 3. Find the value of a.
Answer:
Let f(z) = az³ + 4z² + 3z − 4 and g(z) = z³ − 4z + a. Then f(3) = g(3).
27a + 36 + 9 − 4 = 27 − 12 + a, so 27a + 41 = 15 + a.
26a = −26, so a = −1.

Example 8. Find at least one factor of the polynomial \( p(x) = x^3 - 3x^2 + x + 1 \).
Answer:
The sum of the coefficients is 1 − 3 + 1 + 1 = 0, so p(1) = 0.
Hence (x − 1) is a factor of p(x). Check: p(1) = 1 − 3 + 1 + 1 = 0.

Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

Let α and β be the zeroes of a quadratic polynomial \( f(x) = ax^2 + bx + c \), a ≠ 0. By the factor theorem f(x) can be written as k(x − α)(x − β), where k is a constant. Then \( ax^2 + bx + c = k[x^2 - (\alpha + \beta)x + \alpha\beta] = kx^2 - k(\alpha + \beta)x + k\alpha\beta \). Comparing the coefficients of x², x and the constant terms: a = k, b = −k(α + β) and c = kαβ.

Dividing b and c by a: \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \). Hence

Sum of the zeroes = \( -\dfrac{b}{a} = -\dfrac{\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes = \( \dfrac{c}{a} = \dfrac{\text{constant term}}{\text{coefficient of } x^2} \)

Note: If α and β are the zeroes of a quadratic polynomial f(x), then f(x) = k{x² − (α + β)x + αβ} = k{x² − (sum of the zeroes)x + product of the zeroes}, where k is any non-zero real number.

Relationship Between Zeroes and Coefficients of a Cubic Polynomial

Let α, β, γ be the zeroes of a cubic polynomial \( f(x) = ax^3 + bx^2 + cx + d \), a ≠ 0. By the factor theorem (x − α), (x − β) and (x − γ) are factors, so \( ax^3 + bx^2 + cx + d = k(x-\alpha)(x-\beta)(x-\gamma) = k\{x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma\} \). Comparing coefficients: a = k, b = −k(α + β + γ), c = k(αβ + βγ + γα), d = −kαβγ. Therefore

Sum of the zeroes: \( \alpha + \beta + \gamma = -\dfrac{b}{a} = -\dfrac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \)
Sum of the products of the zeroes taken two at a time: \( \alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a} = \dfrac{\text{coefficient of } x}{\text{coefficient of } x^3} \)
Product of the zeroes: \( \alpha\beta\gamma = -\dfrac{d}{a} = -\dfrac{\text{constant term}}{\text{coefficient of } x^3} \)

Note: A cubic polynomial having α, β, γ as its zeroes is \( f(x) = k\{x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma\} \), where k is any non-zero real number.

Mind it
1. Quartic polynomial: If α, β, γ, δ are the zeroes of \( ax^4 + bx^3 + cx^2 + dx + e \), then \( \alpha+\beta+\gamma+\delta = -\frac{b}{a} \); \( \alpha\beta+\beta\gamma+\gamma\delta+\alpha\gamma+\alpha\delta+\beta\delta = \frac{c}{a} \); \( \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = -\frac{d}{a} \); \( \alpha\beta\gamma\delta = \frac{e}{a} \).

Important Formulae

1. \( (a+b)^2 = a^2 + 2ab + b^2 \)
2. \( (a-b)^2 = a^2 - 2ab + b^2 \)
3. \( a^2 - b^2 = (a+b)(a-b) \)
4. \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \)
5. \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)
6. \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)
7. \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \)
8. \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \)
9. \( a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) \)

Special case: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Examples on Zeroes and Coefficients

Example 1. Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients: (i) \( g(s) = 4s^2 - 4s + 1 \) (ii) \( g(x) = 6x^2 - 3 - 7x \).
Answer:
(i) \( 4s^2 - 4s + 1 = (2s-1)^2 \). Zeroes: 2s − 1 = 0, so \( s = \frac12 \) and \( \frac12 \).
Sum = \( \frac12 + \frac12 = 1 = \frac{-(-4)}{4} = \frac{-\text{coeff. of } s}{\text{coeff. of } s^2} \). Product = \( \frac12 \times \frac12 = \frac14 = \frac{\text{constant term}}{\text{coeff. of } s^2} \). Verified.
(ii) \( 6x^2 - 7x - 3 = (3x+1)(2x-3) \). Zeroes: \( x = -\frac13 \) and \( x = \frac32 \).
Sum = \( -\frac13 + \frac32 = \frac{7}{6} = \frac{-(-7)}{6} \). Product = \( -\frac13 \times \frac32 = -\frac12 = \frac{-3}{6} \). Verified.

Example 2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation: (i) \( -\frac{8}{3}, \frac{4}{3} \) (ii) \( -2\sqrt{3}, -9 \).
Answer:
A quadratic polynomial is f(x) = k[x² − (sum of zeroes)x + product of zeroes].
(i) f(x) = k\( \left(x^2 + \frac83 x + \frac43\right) = \frac{k}{3}(3x^2 + 8x + 4) = \frac{k}{3}(3x+2)(x+2) \). Zeroes: \( -\frac23 \) and −2.
(ii) f(x) = k\( (x^2 + 2\sqrt{3}x - 9) = k(x + 3\sqrt{3})(x - \sqrt{3}) \), because \( x^2 + 3\sqrt3 x - \sqrt3 x - 9 = x(x + 3\sqrt3) - \sqrt3(x + 3\sqrt3) \). Zeroes: \( -3\sqrt{3} \) and \( \sqrt{3} \).

Example 3. Find a quadratic polynomial whose zeroes are 4 and 3 respectively.
Answer:
Sum of zeroes = 4 + 3 = 7 and product = 4 × 3 = 12.
Required polynomial f(x) = k[x² − (sum)x + product] = k[x² − 7x + 12], where k is a non-zero constant.

Example 4. If α and β are the roots (zeroes) of the polynomial \( f(x) = x^2 - 3x + k \) such that α − β = 1, find the value of k.
Answer:
α + β = 3 and αβ = k. Given α − β = 1.
\( (\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta \), so 1 = 9 − 4k, giving 4k = 8.
k = 2

Example 5. If α, β are the zeroes of the polynomial \( f(x) = 2x^2 + 5x + k \) satisfying the relation \( \alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4} \), then find the value of k.
Answer:
α + β = \( -\frac52 \) and αβ = \( \frac k2 \).
\( \alpha^2 + \beta^2 + \alpha\beta = (\alpha+\beta)^2 - \alpha\beta = \frac{25}{4} - \frac{k}{2} = \frac{21}{4} \), so \( \frac k2 = 1 \).
k = 2

Example 6. If both (x − 2) and \( \left(x - \frac12\right) \) are factors of \( px^2 + 5x + r \), then show that p = r.
Answer:
The zeroes are 2 and \( \frac12 \), so the product of the zeroes is \( 2 \times \frac12 = 1 = \frac{r}{p} \).
Hence r = p.

Example 7. Find the zeroes of the quadratic polynomial \( 9x^2 - 5 \) and verify the relation between the zeroes and its coefficients.
Answer:
\( 9x^2 - 5 = (3x - \sqrt5)(3x + \sqrt5) \). Zeroes: \( \frac{\sqrt5}{3} \) and \( -\frac{\sqrt5}{3} \).
Sum = 0 = \( \frac{-0}{9} \). Product = \( -\frac{5}{9} = \frac{\text{constant term}}{\text{coefficient of } x^2} \). Verified.

Example 8. If α and β are the zeroes of the quadratic polynomial \( f(x) = ax^2 + bx + c \), then calculate the value of \( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} \).
Answer:
α + β = \( -\frac ba \), αβ = \( \frac ca \).
\( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3+\beta^3}{\alpha\beta} = \frac{(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)}{\alpha\beta} = \frac{-\frac{b^3}{a^3} + \frac{3bc}{a^2}}{\frac ca} = \frac{3abc - b^3}{a^2 c} \).

Example 9. If α and β are the zeroes of the quadratic polynomial \( p(s) = 3s^2 - 6s + 4 \), find the value of \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac1\alpha + \frac1\beta\right) + 3\alpha\beta \).
Answer:
α + β = 2 and αβ = \( \frac43 \).
\( \frac{\alpha^2+\beta^2}{\alpha\beta} + \frac{2(\alpha+\beta)}{\alpha\beta} + 3\alpha\beta = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} + \frac{2(\alpha+\beta)}{\alpha\beta} + 3\alpha\beta \)
\( = \frac{4 - \frac83}{\frac43} + \frac{4}{\frac43} + 4 = 1 + 3 + 4 = 8 \).
8

Example 10. Verify that the numbers given alongside the polynomial \( x^4 + 2x^3 - 7x^2 - 8x + 12 \) are its zeroes: −3, −2, 1, 2.
Answer:
Let p(x) = x⁴ + 2x³ − 7x² − 8x + 12.
p(−3) = 81 − 54 − 63 + 24 + 12 = 0; p(−2) = 16 − 16 − 28 + 16 + 12 = 0; p(1) = 1 + 2 − 7 − 8 + 12 = 0; p(2) = 16 + 16 − 28 − 16 + 12 = 0.
So −3, −2, 1 and 2 are zeroes of p(x).

Example 11. If α and β are the zeroes of the polynomial \( ax^2 + bx + c \), find a polynomial whose zeroes are \( \frac{1}{a\alpha + b} \) and \( \frac{1}{a\beta + b} \).
Answer:
α + β = \( -\frac ba \) and αβ = \( \frac ca \).
Sum of the new zeroes = \( \frac{a\beta + b + a\alpha + b}{(a\alpha+b)(a\beta+b)} = \frac{a(\alpha+\beta) + 2b}{a^2\alpha\beta + ab(\alpha+\beta) + b^2} = \frac{-b + 2b}{ac - b^2 + b^2} = \frac{b}{ac} \).
Product of the new zeroes = \( \frac{1}{a^2\alpha\beta + ab(\alpha+\beta) + b^2} = \frac{1}{ac} \).
Required polynomial: \( x^2 - \frac{b}{ac}x + \frac{1}{ac} \).

Division Algorithm for Polynomials

If f(x) and p(x) are any two polynomials with p(x) ≠ 0, then we can find polynomials q(x) and r(x) such that f(x) = p(x) × q(x) + r(x), that is, dividend = (divisor × quotient) + remainder, where r(x) = 0 or degree of r(x) < degree of p(x).

1. If r(x) = 0, then p(x) is a factor of f(x).
2. If deg f(x) > deg p(x), then deg q(x) = deg f(x) − deg p(x).
3. If deg f(x) = deg p(x), then deg q(x) = 0 and deg r(x) < deg p(x).

Working Rule to Divide a Polynomial by Another Polynomial

1. Arrange the terms of the dividend and the divisor in decreasing order of their degrees.
2. To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest degree term of the divisor.
3. To obtain the second term of the quotient, divide the highest degree term of the new dividend (the remainder so far) by the highest degree term of the divisor.
4. Continue this process till the degree of the remainder is less than the degree of the divisor.

Examples on the Division Algorithm

Example 1. Check whether the first polynomial is a factor of the second by applying the division algorithm: \( x^2 + 3x + 1 \), \( 3x^4 + 5x^3 - 7x^2 + 2x + 2 \).
Answer:
Dividing \( 3x^4 + 5x^3 - 7x^2 + 2x + 2 \) by \( x^2 + 3x + 1 \): first term 3x² (subtract \( 3x^4 + 9x^3 + 3x^2 \) leaving \( -4x^3 - 10x^2 + 2x + 2 \)), next term −4x (subtract \( -4x^3 - 12x^2 - 4x \) leaving \( 2x^2 + 6x + 2 \)), next term 2 (subtract \( 2x^2 + 6x + 2 \) leaving 0).
Quotient = \( 3x^2 - 4x + 2 \), remainder = 0. Since the remainder is zero, \( x^2 + 3x + 1 \) is a factor.
Check: \( (3x^2 - 4x + 2)(x^2 + 3x + 1) = 3x^4 + 5x^3 - 7x^2 + 2x + 2 \), the dividend.

Example 2. Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x), where \( p(x) = x^4 - 3x^2 + 4x + 5 \) and \( g(x) = x^2 + 1 - x \).
Answer:
Write g(x) = x² − x + 1. Divide: first term x² (subtract \( x^4 - x^3 + x^2 \) leaving \( x^3 - 4x^2 + 4x + 5 \)), next term x (subtract \( x^3 - x^2 + x \) leaving \( -3x^2 + 3x + 5 \)), next term −3 (subtract \( -3x^2 + 3x - 3 \) leaving 8).
We stop because the degree of 8 is less than the degree of g(x). Quotient = x² + x − 3, remainder = 8.
Check: \( (x^2 + x - 3)(x^2 - x + 1) + 8 = x^4 - 3x^2 + 4x - 3 + 8 = x^4 - 3x^2 + 4x + 5 \). The division algorithm is verified.

Example 3. If the polynomial \( p(x) = 2x^5 - 15x^4 + 40x^3 - 34x^2 + ax + b \) is divisible by \( g(x) = x^2 - 4x + 6 \), then find the value of a + b.
Answer:
Dividing p(x) by g(x), the quotient is \( 2x^3 - 7x^2 + 8 \) and the remainder is (32 + a)x + (b − 48).
As p(x) is divisible by g(x), the remainder is zero: 32 + a = 0 and b − 48 = 0, so a = −32 and b = 48.
a + b = −32 + 48 = 16.

Example 4. If x + y + z = 1, x² + y² + z² = 21 and xyz = 8, then find the value of (1 − x)(1 − y)(1 − z).
Answer:
Squaring x + y + z = 1: x² + y² + z² + 2(xy + yz + zx) = 1, so 21 + 2(xy + yz + zx) = 1 and xy + yz + zx = −10.
(1 − x)(1 − y)(1 − z) = 1 − (x + y + z) + (xy + yz + zx) − xyz = 1 − 1 − 10 − 8 = −18.

Example 5. Find the value of b for which the polynomial \( 2x^3 + 9x^2 - x - b \) is exactly divisible by 2x + 3.
Answer:
Dividing by 2x + 3: quotient \( x^2 + 3x - 5 \), remainder −b + 15.
For exact divisibility the remainder is zero: −b + 15 = 0, so b = 15.

Example 6. Find the value that must be subtracted from the expression \( 8x^4 + 14x^3 - 2x^2 + 8x - 12 \) so that it is exactly divisible by \( 4x^2 + 3x - 2 \).
Answer:
Dividing, the quotient is \( 2x^2 + 2x - 1 \) and the remainder is 15x − 14.
For exact divisibility the remainder must be zero, so the expression that must be subtracted is 15x − 14.

Example 7. If \( x^n - py^n + qz^n \) is exactly divisible by \( x^2 - (ay + bz)x + abyz \), then find the value of \( \frac{p}{a^n} - \frac{q}{b^n} \).
Answer:
Factorising, x² − (ay + bz)x + abyz = (x − ay)(x − bz). Let f(x) = xⁿ − pyⁿ + qzⁿ. By the factor theorem f(ay) = 0 and f(bz) = 0.
f(ay): aⁿyⁿ − pyⁿ + qzⁿ = 0 ... (i) f(bz): bⁿzⁿ − pyⁿ + qzⁿ = 0 ... (ii)
Subtracting, \( a^n y^n = b^n z^n \), so \( y^n = \frac{b^n z^n}{a^n} \). Substituting in (ii): \( b^n z^n - p\frac{b^n z^n}{a^n} + qz^n = 0 \).
Dividing by \( b^n z^n \): \( 1 - \frac{p}{a^n} + \frac{q}{b^n} = 0 \), so \( \frac{p}{a^n} - \frac{q}{b^n} = 1 \).

Example 8. Let a, b, c, x, y and z be numbers such that \( a = \frac{b+c}{x-2},\ b = \frac{c+a}{y-2},\ c = \frac{a+b}{z-2} \). If xy + yz + zx = 67 and x + y + z = 2010, find the value of xyz + 5900.
Answer:
From the first relation x − 2 = (b + c)/a, so x − 1 = (a + b + c)/a and 1/(x − 1) = a/(a + b + c). Similarly 1/(y − 1) = b/(a + b + c) and 1/(z − 1) = c/(a + b + c).
Adding: 1/(x − 1) + 1/(y − 1) + 1/(z − 1) = 1, so (y − 1)(z − 1) + (x − 1)(z − 1) + (x − 1)(y − 1) = (x − 1)(y − 1)(z − 1).
Left side = (xy + yz + zx) − 2(x + y + z) + 3. Right side = xyz − (xy + yz + zx) + (x + y + z) − 1.
So xyz = 2(xy + yz + zx) − 3(x + y + z) + 4 = 2(67) − 3(2010) + 4 = 134 − 6030 + 4 = −5892.
xyz + 5900 = −5892 + 5900 = 8.

Summary

1. An algebraic expression \( f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \) is a polynomial, where \( a_0, \ldots, a_n \) are real numbers and all the powers of x are non-negative integers.
2. The degree of a polynomial is the highest exponent of the variable. A zero of a polynomial is a number for which the value of the polynomial is zero; the zeroes are the points where the graph meets the x-axis.
3. f(x) = ax + b (a ≠ 0) is a linear polynomial and its graph is a straight line.
4. f(x) = ax² + bx + c (a ≠ 0) is a quadratic polynomial and its graph is a parabola, opening upwards if a is positive and downwards if a is negative.
5. For a quadratic polynomial: sum of the zeroes α + β = \( -\frac ba \); product of the zeroes αβ = \( \frac ca \). A quadratic polynomial is x² − (sum of zeroes)x + (product of zeroes).
6. For a cubic polynomial \( ax^3 + bx^2 + cx + d \) with zeroes α, β, γ: α + β + γ = \( -\frac ba \); αβ + βγ + γα = \( \frac ca \); αβγ = \( -\frac da \).
7. Division algorithm: dividend = divisor × quotient + remainder, that is, f(x) = g(x) × q(x) + r(x), where r(x) = 0 or deg r(x) < deg g(x).