Polynomials Notes

2

Polynomials

Detailed Theory • Easy English • Board Exam Focus

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Chapter Roadmap

1. Basic Terms 2. Polynomial in One Variable 3. Degree and Types 4. Zeroes of a Polynomial 5. Geometrical Meaning of Zeroes 6. Quadratic Polynomial 7. Zeroes and Coefficients 8. Cubic Polynomial 9. Board-Style Questions 10. Final Revision

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1. Constants, Variables and Algebraic Expressions

Constant

A constant is a number whose value does not change.

Examples

\(2,\;5,\;-3,\;100,\;\pi\) are constants.

Variable

A variable is a quantity whose value can change. We usually use letters such as \(x,y,z\).

Example

In \(3x+5\), \(x\) is the variable, \(3\) is the coefficient of \(x\), and \(5\) is the constant.

Algebraic Expression

An algebraic expression is a mathematical expression made using constants, variables and operations such as addition, subtraction and multiplication.

\[\boxed{3x^2-5x+7}\]

Terms

The parts of an algebraic expression separated by \(+\) or \(-\) signs are called its terms.

In \(3x^2-5x+7\), the terms are \(3x^2,\;-5x,\;7\).

Therefore, it has 3 terms.

Solved Example 1 — Identify the parts

Question: Identify the variable, coefficient and constant in \(5x+7\).

1. Look at the expression: \(5x+7\).
2. The letter is \(x\), so \(x\) is the variable.
3. The number multiplied by \(x\) is \(5\), so \(5\) is the coefficient.
4. The number without a variable is \(7\), so \(7\) is the constant.
5. Answer: Variable \(=x\), coefficient \(=5\), constant \(=7\).

Solved Example 2 — Count the terms

Question: How many terms are there in \(3x^2-4x+9\)?

1. Write the expression: \(3x^2-4x+9\).
2. Separate the terms using the \(+\) and \(-\) signs.
3. The terms are \(3x^2,\;-4x,\;9\).
4. Count them: \(1+1+1=3\).
5. Answer: There are \(\boxed{3}\) terms.

Solved Example 3 — Read an expression

Question: Identify the terms and coefficient of \(x\) in \(7x^2+3x-5\).

1. Write the expression: \(7x^2+3x-5\).
2. The terms are \(7x^2,\;3x,\;-5\).
3. The term containing \(x\) is \(3x\).
4. The coefficient of \(x\) is \(3\).
5. Answer: Terms \(=7x^2,3x,-5\); coefficient of \(x=\boxed{3}\).

2. Polynomial in One Variable

Definition

A polynomial in one variable \(x\) is an expression of the form

\[ p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0 \]

where \(a_0,a_1,\ldots,a_n\) are real numbers and \(n\) is a non-negative integer.

Examples of Polynomials

• \(2x+3\)
• \(x^2-5x+6\)
• \(3x^3+2x^2-x+4\)
• \(7\)

Expressions That Are Not Polynomials

The powers of the variable in a polynomial must be non-negative integers.

Not polynomials: \(\frac{1}{x}\), \(\sqrt{x}\), \(x^{-2}\).

Reason: their powers of \(x\) are \(-1,\frac12,-2\), which are not non-negative integers.

Solved Example 1 — Is it a polynomial?

Question: Check whether \(4x^2+3x-7\) is a polynomial in \(x\).

1. Check the powers of \(x\).
2. The powers are \(2,1,0\).
3. All these powers are non-negative integers.
4. Therefore, the expression satisfies the basic condition for a polynomial.
5. Answer: Yes, \(4x^2+3x-7\) is a polynomial in \(x\).

Solved Example 2 — Not a polynomial

Question: Check whether \(\frac{3}{x}+2\) is a polynomial.

1. Rewrite \(\frac{3}{x}\) as \(3x^{-1}\).
2. The power of \(x\) is \(-1\).
3. A polynomial cannot have a negative integer power of the variable.
4. Therefore, the expression does not satisfy the condition.
5. Answer: \(\boxed{\frac{3}{x}+2}\) is not a polynomial.

Solved Example 3 — Write in standard form

Question: Write \(6+2x^3-5x\) in descending powers of \(x\).

1. Look at the powers: \(3,1,0\).
2. Arrange them from highest to lowest: \(3,1,0\).
3. Write the terms in that order.
4. So, \(6+2x^3-5x=2x^3-5x+6\).
5. Answer: \(\boxed{2x^3-5x+6}\).

3. Degree and Types of Polynomials

Degree of a Polynomial

The highest power of the variable with a non-zero coefficient is called the degree of the polynomial.

Polynomial	Degree	Type
\(7\)	0	Constant
\(2x+5\)	1	Linear
\(x^2-3x+2\)	2	Quadratic
\(x^3+2x-1\)	3	Cubic
\(x^4+x^2+1\)	4	Quartic

Types Based on Degree

Linear polynomial: degree \(1\), general form \(ax+b,\;a\neq0\).

Quadratic polynomial: degree \(2\), general form \(ax^2+bx+c,\;a\neq0\).

Cubic polynomial: degree \(3\), general form \(ax^3+bx^2+cx+d,\;a\neq0\).

Types Based on Number of Terms

• Monomial: one term, e.g. \(5x^2\).
• Binomial: two terms, e.g. \(x+3\).
• Trinomial: three terms, e.g. \(x^2+5x+6\).

Solved Example 1 — Find degree

Question: Find the degree of \(5x^4-2x^2+x-8\).

1. Write all powers of \(x\): \(4,2,1,0\).
2. Find the highest power.
3. The highest power is \(4\).
4. Therefore, the degree is \(4\).
5. Answer: \(\boxed{4}\).

Solved Example 2 — Identify type

Question: Identify the type of \(3x^2-7x+2\) by degree.

1. The highest power of \(x\) is \(2\).
2. Therefore, its degree is \(2\).
3. A polynomial of degree \(2\) is called a quadratic polynomial.
4. Answer: It is a \(\boxed{\text{quadratic polynomial}}\).

Solved Example 3 — Type by terms

Question: Classify \(9x^2-4\) according to the number of terms.

1. The expression is \(9x^2-4\).
2. The terms are \(9x^2\) and \(-4\).
3. There are \(2\) terms.
4. A polynomial with two terms is called a binomial.
5. Answer: \(\boxed{\text{Binomial}}\).

4. Zeroes of a Polynomial

Meaning of Zero

A number \(k\) is called a zero of a polynomial \(p(x)\) if

\[\boxed{p(k)=0}\]

Example 1: Find the zero of \(p(x)=2x-6\)

1. Put \(p(x)=0\).
2. \(2x-6=0\)
3. \(2x=6\)
4. \(x=3\)
5. Therefore, the zero is \(\boxed{3}\).

Example 2: Check whether \(2\) is a zero of \(p(x)=x^2-4\)

1. Given \(p(x)=x^2-4\).
2. Put \(x=2\).
3. \(p(2)=2^2-4\)
4. \(=4-4=0\)
5. Since \(p(2)=0\), \(2\) is a zero of the polynomial.

Board Tip: To check whether a given number is a zero, simply substitute that number for \(x\). If the answer is \(0\), it is a zero.

Solved Example 1 — Linear zero

Question: Find the zero of \(p(x)=5x-15\).

1. For a zero, put \(p(x)=0\).
2. \(5x-15=0\).
3. Add \(15\) to both sides: \(5x=15\).
4. Divide both sides by \(5\): \(x=3\).
5. Answer: The zero is \(\boxed{3}\).

Solved Example 2 — Check a zero

Question: Check whether \(3\) is a zero of \(p(x)=x^2-9\).

1. Put \(x=3\).
2. \(p(3)=3^2-9\).
3. \(=9-9\).
4. \(=0\).
5. Since \(p(3)=0\), \(\boxed{3}\) is a zero.

Solved Example 3 — Two zeroes

Question: Find the zeroes of \(x^2-5x+6\).

1. Set the polynomial equal to zero: \(x^2-5x+6=0\).
2. Find two numbers whose product is \(6\) and sum is \(-5\): \(-2,-3\).
3. So, \(x^2-5x+6=(x-2)(x-3)\).
4. Therefore, \((x-2)(x-3)=0\).
5. Thus \(x=2\) or \(x=3\).
6. Answer: Zeroes are \(\boxed{2,3}\).

5. Geometrical Meaning of the Zeroes

The graph of a polynomial \(y=p(x)\) can meet or touch the \(x\)-axis at certain points.

At every point where the graph intersects or touches the \(x\)-axis,

\[y=0\]

Therefore, the corresponding \(x\)-coordinate is a zero of the polynomial.

Important Result

The zeroes of a polynomial are the \(x\)-coordinates of the points where its graph intersects or touches the \(x\)-axis.

Remember for Board Exam:

• A linear polynomial has exactly one zero.
• A quadratic polynomial can have at most two zeroes.
• A cubic polynomial can have at most three zeroes.

Solved Example 1 — Read a graph

Question: If a graph cuts the \(x\)-axis at \(x=2\), what is the zero?

1. On the \(x\)-axis, \(y=0\).
2. At the cutting point, the corresponding \(x\)-coordinate is \(2\).
3. Therefore, \(p(2)=0\).
4. Hence, \(2\) is a zero.
5. Answer: \(\boxed{2}\).

Solved Example 2 — Number of zeroes

Question: A quadratic graph intersects the \(x\)-axis at \(x=-1\) and \(x=4\). Find its zeroes.

1. Zeroes are the \(x\)-coordinates where the graph meets the \(x\)-axis.
2. The first point has \(x=-1\).
3. The second point has \(x=4\).
4. Therefore, the two zeroes are \(-1\) and \(4\).
5. Answer: \(\boxed{-1,4}\).

Solved Example 3 — No intersection

Question: What can we say if the graph of a polynomial never meets the \(x\)-axis?

1. A zero occurs where \(y=0\).
2. The \(x\)-axis represents \(y=0\).
3. If the graph never meets or touches the \(x\)-axis, there is no point with \(y=0\).
4. Therefore, the polynomial has no real zero.
5. Answer: It has \(\boxed{0}\) real zeroes.

6. Quadratic Polynomial and Its Zeroes

The general form of a quadratic polynomial is

\[\boxed{p(x)=ax^2+bx+c,\quad a\neq0}\]

If its zeroes are \(\alpha\) and \(\beta\), then

\[ \boxed{\alpha+\beta=-\frac{b}{a}} \qquad \boxed{\alpha\beta=\frac{c}{a}} \]

Example: \(x^2-5x+6\)

1. Compare \(x^2-5x+6\) with \(ax^2+bx+c\).
2. So, \(a=1,\;b=-5,\;c=6\).
3. Sum of zeroes: \[ \alpha+\beta=-\frac{b}{a}=-\frac{-5}{1}=5 \]
4. Product of zeroes: \[ \alpha\beta=\frac{c}{a}=\frac61=6 \]
5. Indeed, the zeroes are \(2\) and \(3\), because \(2+3=5\) and \(2\times3=6\).

Solved Example 1 — Identify \(a,b,c\)

Question: Find \(a,b,c\) for \(3x^2-8x+5\).

1. Compare with \(ax^2+bx+c\).
2. Coefficient of \(x^2\): \(a=3\).
3. Coefficient of \(x\): \(b=-8\).
4. Constant term: \(c=5\).
5. Answer: \(\boxed{a=3,\ b=-8,\ c=5}\).

Solved Example 2 — Sum and product

Question: Find the sum and product of zeroes of \(2x^2-7x+3\).

1. Compare with \(ax^2+bx+c\).
2. So \(a=2,\;b=-7,\;c=3\).
3. Sum \(=-\frac ba=-\frac{-7}{2}=\frac72\).
4. Product \(=\frac ca=\frac32\).
5. Answer: Sum \(=\boxed{\frac72}\), product \(=\boxed{\frac32}\).

Solved Example 3 — Verify by factorisation

Question: Verify the zeroes of \(x^2-7x+12\) using the coefficient relations.

1. Factorise: \(x^2-7x+12=(x-3)(x-4)\).
2. Therefore, \(\alpha=3,\;\beta=4\).
3. Sum \(=3+4=7\).
4. Using the formula, \(-b/a=-(-7)/1=7\).
5. Product \(=3\times4=12\).
6. Using the formula, \(c/a=12/1=12\).
7. Answer: Both relationships are verified.

7. Relationship Between Zeroes and Coefficients

For a Quadratic Polynomial

Let

\[p(x)=ax^2+bx+c\]

and let its zeroes be \(\alpha,\beta\).

\[ \boxed{\alpha+\beta=-\frac ba} \] \[ \boxed{\alpha\beta=\frac ca} \]

Forming a Quadratic Polynomial from Given Zeroes

If \(\alpha\) and \(\beta\) are given zeroes, then a quadratic polynomial with these zeroes is

\[ \boxed{x^2-(\alpha+\beta)x+\alpha\beta} \]

Example: Form a polynomial whose zeroes are \(3\) and \(-2\)

1. Given \(\alpha=3,\;\beta=-2\).
2. Find their sum: \[ \alpha+\beta=3+(-2)=1 \]
3. Find their product: \[ \alpha\beta=3(-2)=-6 \]
4. Use the formula: \[ x^2-(\alpha+\beta)x+\alpha\beta \]
5. Substitute the values: \[ x^2-(1)x-6 \]
6. Therefore, \[ \boxed{x^2-x-6} \] is a required polynomial.

Finding a Polynomial When Zeroes and a Constant Multiple Are Allowed

Any non-zero constant multiple of a polynomial has the same zeroes.

\[ k[x^2-(\alpha+\beta)x+\alpha\beta],\quad k\neq0 \]

Solved Example 1 — Find unknown coefficient

Question: If one zero of \(2x^2+kx+3\) is \(1\), find \(k\).

1. Since \(1\) is a zero, \(p(1)=0\).
2. \(2(1)^2+k(1)+3=0\).
3. \(2+k+3=0\).
4. \(k+5=0\).
5. \(\boxed{k=-5}\).

Solved Example 2 — Form polynomial

Question: Form a quadratic polynomial whose zeroes are \(4\) and \(1\).

1. Let \(\alpha=4,\;\beta=1\).
2. Sum: \(\alpha+\beta=4+1=5\).
3. Product: \(\alpha\beta=4\times1=4\).
4. Use \(x^2-(\alpha+\beta)x+\alpha\beta\).
5. Substitute: \(x^2-5x+4\).
6. Answer: \(\boxed{x^2-5x+4}\).

Solved Example 3 — Expression in zeroes

Question: If \(\alpha,\beta\) are zeroes of \(x^2-6x+5\), find \(\alpha^2+\beta^2\).

1. Here \(a=1,\;b=-6,\;c=5\).
2. \(\alpha+\beta=-b/a=6\).
3. \(\alpha\beta=c/a=5\).
4. Use \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\).
5. \(=6^2-2(5)=36-10=26\).
6. Answer: \(\boxed{26}\).

8. Cubic Polynomial

The general form of a cubic polynomial is

\[\boxed{p(x)=ax^3+bx^2+cx+d,\quad a\neq0}\]

If its zeroes are \(\alpha,\beta,\gamma\), then

\[ \boxed{\alpha+\beta+\gamma=-\frac ba} \] \[ \boxed{\alpha\beta+\beta\gamma+\gamma\alpha=\frac ca} \] \[ \boxed{\alpha\beta\gamma=-\frac da} \]

Example: \(2x^3-5x^2-x+6\)

Here \(a=2,\;b=-5,\;c=-1,\;d=6\).

\[ \alpha+\beta+\gamma=-\frac{-5}{2}=\frac52 \] \[ \alpha\beta+\beta\gamma+\gamma\alpha=\frac{-1}{2}=-\frac12 \] \[ \alpha\beta\gamma=-\frac62=-3 \]

Solved Example 1 — Identify coefficients

Question: Find \(a,b,c,d\) for \(3x^3-2x^2+5x-7\).

1. Compare with \(ax^3+bx^2+cx+d\).
2. Coefficient of \(x^3\): \(a=3\).
3. Coefficient of \(x^2\): \(b=-2\).
4. Coefficient of \(x\): \(c=5\).
5. Constant: \(d=-7\).
6. Answer: \(\boxed{a=3,b=-2,c=5,d=-7}\).

Solved Example 2 — Sum of three zeroes

Question: For \(2x^3-9x^2+4x+1\), find the sum of zeroes.

1. Compare with \(ax^3+bx^2+cx+d\).
2. So \(a=2,\;b=-9\).
3. For zeroes \(\alpha,\beta,\gamma\), \(\alpha+\beta+\gamma=-b/a\).
4. \(=-(-9)/2=9/2\).
5. Answer: \(\boxed{\frac92}\).

Solved Example 3 — Product of three zeroes

Question: For \(x^3-6x^2+11x-6\), find \(\alpha\beta\gamma\).

1. Compare with \(ax^3+bx^2+cx+d\).
2. Here \(a=1,\;d=-6\).
3. Use \(\alpha\beta\gamma=-d/a\).
4. \(=-(-6)/1=6\).
5. Answer: \(\boxed{6}\).

9. Important Board-Exam Question Types

Type 1: Find the zeroes of a polynomial.

Type 2: Verify the relationship between zeroes and coefficients.

Type 3: Find unknown coefficients when zeroes are given.

Type 4: Form a quadratic polynomial when its zeroes are given.

Type 5: Find the zeroes from a graph.

Type 6: Use the sum/product of zeroes to find an expression involving zeroes.

10. 10 Fully Solved Board-Style Questions

Question 1. Find the zero of \(p(x)=3x-12\).

Solution:

For the zero,

\[p(x)=0\]

Therefore,

\[3x-12=0\]

\[3x=12\]

\[\boxed{x=4}\]

Answer: The zero is \(\boxed{4}\).

Question 2. Find the zeroes of \(x^2-7x+12\).

Solution:

We factorise the polynomial:

\[ x^2-7x+12=x^2-3x-4x+12 \]

Taking common factors,

\[ =x(x-3)-4(x-3) \]

\[ =(x-3)(x-4) \]

For zeroes:

\[ (x-3)(x-4)=0 \]

Therefore,

\[ x-3=0\Rightarrow x=3 \] \[ x-4=0\Rightarrow x=4 \]

Answer: The zeroes are \(\boxed{3,4}\).

Question 3. Verify the relationship for \(2x^2-5x+2\).

Solution:

Factorise:

\[ 2x^2-5x+2=(2x-1)(x-2) \]

Therefore, zeroes are

\[ \alpha=\frac12,\qquad\beta=2 \]

Sum:

\[ \alpha+\beta=\frac12+2=\frac52 \]

Using \(a=2,b=-5\):

\[ -\frac ba=-\frac{-5}{2}=\frac52 \]

Hence, \(\alpha+\beta=-b/a\).

Product:

\[ \alpha\beta=\frac12(2)=1 \]

Using \(c=2,a=2\):

\[ \frac ca=\frac22=1 \]

Hence, \(\alpha\beta=c/a\).

Verified.

Question 4. Find a quadratic polynomial whose zeroes are \(5\) and \(-3\).

Solution:

\(\alpha=5,\;\beta=-3\)

\[ \alpha+\beta=5-3=2 \]

\[ \alpha\beta=5(-3)=-15 \]

Required polynomial:

\[ x^2-(\alpha+\beta)x+\alpha\beta \]

\[ =x^2-2x-15 \]

Answer: \(\boxed{x^2-2x-15}\).

Question 5. If the zeroes of \(x^2-(k+2)x+2k\) are equal, find \(k\).

Solution:

For equal zeroes, the discriminant must be zero.

\[ D=b^2-4ac=0 \]

Here \(a=1,\;b=-(k+2),\;c=2k\).

\[ (k+2)^2-4(1)(2k)=0 \]

\[ k^2+4k+4-8k=0 \]

\[ k^2-4k+4=0 \]

\[ (k-2)^2=0 \]

Hence,

\[\boxed{k=2}\]

Question 6. If one zero of \(2x^2+px+6\) is \(2\), find \(p\).

Solution:

Since \(2\) is a zero,

\[ p(2)=0 \]

\[ 2(2)^2+p(2)+6=0 \]

\[ 8+2p+6=0 \]

\[ 14+2p=0 \]

\[ 2p=-14 \]

\[ \boxed{p=-7} \]

Question 7. If the sum of zeroes is \(6\) and their product is \(8\), form the polynomial.

Solution:

Given:

\[ \alpha+\beta=6,\qquad\alpha\beta=8 \]

Required polynomial:

\[ x^2-(\alpha+\beta)x+\alpha\beta \]

\[ =x^2-6x+8 \]

Answer: \(\boxed{x^2-6x+8}\).

Question 8. Find the zeroes of \(4x^2-4x-3\).

Solution:

\[ 4x^2-4x-3=0 \]

Using the quadratic formula:

\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

Here \(a=4,b=-4,c=-3\).

\[ x=\frac{4\pm\sqrt{(-4)^2-4(4)(-3)}}{8} \]

\[ =\frac{4\pm\sqrt{16+48}}8 =\frac{4\pm8}{8} \]

Therefore,

\[ x=\frac{12}{8}=\frac32 \quad\text{or}\quad x=\frac{-4}{8}=-\frac12 \]

Answer: \(\boxed{\frac32,-\frac12}\).

Question 9. If \(\alpha,\beta\) are zeroes of \(x^2-8x+12\), find \(\alpha^2+\beta^2\).

Solution:

For the given polynomial,

\[ \alpha+\beta=8,\qquad\alpha\beta=12 \]

We know:

\[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta \]

Substitute the values:

\[ =8^2-2(12) \]

\[ =64-24=40 \]

Answer: \(\boxed{40}\).

Question 10. If the zeroes of \(x^2-5x+k\) are reciprocal to each other, find \(k\).

Solution:

If the zeroes are reciprocal, their product is \(1\).

\[ \alpha\beta=1 \]

For \(x^2-5x+k\),

\[ \alpha\beta=\frac ca=\frac{k}{1}=k \]

Therefore,

\[ k=1 \]

Answer: \(\boxed{k=1}\).

11. How to Write Polynomials Answers in the Board Exam

1. Write the given polynomial clearly.
2. Compare it with the standard form \(ax^2+bx+c\) whenever required.
3. Write the values of \(a,b,c\).
4. State the formula before substituting values.
5. Show every important algebraic step.
6. Write the final answer in a box.
7. For verification questions, calculate both sides separately.
8. Do not skip factorisation steps in a long question.

Common Mistakes to Avoid

• Forgetting the negative sign in \(-b/a\).
• Using \(c/a\) incorrectly for the product of zeroes.
• Calling \(\sqrt{x}\) a polynomial in \(x\).
• Not comparing the polynomial correctly with \(ax^2+bx+c\).
• Writing only the final answer without working in a step-marking question.

12. Final 96% Target Revision

Must-Remember Formulas

\[ \boxed{\text{Zero of }p(x)=k\iff p(k)=0} \]

\[ \boxed{\alpha+\beta=-\frac ba} \qquad \boxed{\alpha\beta=\frac ca} \]

\[ \boxed{x^2-(\alpha+\beta)x+\alpha\beta} \]

\[ \boxed{\alpha+\beta+\gamma=-\frac ba} \]

\[ \boxed{\alpha\beta+\beta\gamma+\gamma\alpha=\frac ca} \]

\[ \boxed{\alpha\beta\gamma=-\frac da} \]

One-Minute Checklist

• ✓ Know the definition of polynomial.
• ✓ Know what makes an expression a polynomial.
• ✓ Identify degree correctly.
• ✓ Find zeroes by substitution or factorisation.
• ✓ Understand the graph meaning of zeroes.
• ✓ Memorise the quadratic zero-coefficient relations.
• ✓ Practise forming a polynomial from given zeroes.
• ✓ Show complete steps in board answers.

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Class 10 Maths • Chapter 2: Polynomials

Designed as easy-to-read handwritten-style board exam notes.