How To Form A Polynomial With The Given Zeroes

Form A Polynomial With The Given Zeros

Let zeros of a quadratic polynomial be α and β.
x = β,               x = β
x – α = 0,   x ­– β = 0
The obviously the quadratic polynomial is
(x – α) (x – β)
i.e.,  x2 – (α + β) x + αβ
x2 – (Sum of the zeros)x + Product of the zeros

Form A Polynomial With The Given Zeros Example Problems With Solutions

Example 1:    Form the quadratic polynomial whose zeros are 4 and 6.
Sol.    Sum of the zeros = 4 + 6 = 10
Product of the zeros = 4 × 6 = 24
Hence the polynomial formed
= x2 – (sum of zeros) x + Product of zeros
= x2 – 10x + 24

Example 2:    Form the quadratic polynomial whose zeros are –3, 5.
Sol.    Here, zeros are – 3 and 5.
Sum of the zeros = – 3 + 5 = 2
Product of the zeros = (–3) × 5 = – 15
Hence the polynomial formed
= x2 – (sum of zeros) x + Product of zeros
= x2 – 2x – 15

Example 3:    Find a quadratic polynomial whose sum of zeros and product of zeros are respectively \(\frac { 1 }{ 2 }\), – 1
Sol.   Let the polynomial be ax2 + bx + c and its zeros be  α and β.
(i) Here, α + β = \(\frac { 1 }{ 4 }\) and α.β = – 1
Thus the polynomial formed
= x2 – (Sum of zeros) x + Product of zeros
\(={{\text{x}}^{\text{2}}}-\left( \frac{1}{4} \right)\text{x}-1={{\text{x}}^{\text{2}}}-\frac{\text{x}}{\text{4}}-1\)
The other polynomial are   \(\text{k}\left( {{\text{x}}^{\text{2}}}\text{-}\frac{\text{x}}{\text{4}}\text{-1} \right)\)
If k = 4, then the polynomial is 4x2 – x – 4.

Example 4:    Find a quadratic polynomial whose sum of zeros and product of zeros are respectively \(\sqrt { 2 }\),  \(\frac { 1 }{ 3 }\)
Sol. Here, α + β =\(\sqrt { 2 }\), αβ = \(\frac { 1 }{ 3 }\)
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – \(\sqrt { 2 }\) x + \(\frac { 1 }{ 3 }\)
Other polynomial are   \(\text{k}\left( {{\text{x}}^{\text{2}}}\text{-}\frac{\text{x}}{\text{3}}\text{-1} \right)\)
If k = 3, then the polynomial is
3x2 – \(3\sqrt { 2 }x\)  + 1

Example 5:    Find a quadratic polynomial whose sum of zeros and product of zeros are respectively 0, √5
Sol. Here, α + β = 0, αβ = √5
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – (0) x + √5 = x2 + √5

Example 6:    Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2, – 7 and –14, respectively.
Sol.    Let the cubic polynomial be ax3 + bx2 + cx + d
⇒ x3 + \(\frac { b }{ a }\)x2 + \(\frac { c }{ a }\)x + \(\frac { d }{ a }\) …(1)
and its zeroes are α, β and γ then
α + β + γ = 2 = \(\frac { -b }{ a }\)
αβ + βγ + γα = – 7 = \(\frac { c }{ a }\)
αβγ = – 14 = \(\frac { -d }{ a }\)
Putting the values of   \(\frac { b }{ a }\), \(\frac { c }{ a }\),  and \(\frac { d }{ a }\)  in (1), we get
x3 + (–2) x2 + (–7)x + 14
⇒ x3 – 2x2 – 7x + 14

Example 7:   Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively.
Sol.    Let the cubic polynomial be ax3 + bx2 + cx + d
⇒ x3 + \(\frac { b }{ a }\)x2 + \(\frac { c }{ a }\)x + \(\frac { d }{ a }\) …(1)
and its zeroes are α, β and γ then
α + β + γ = 0 = \(\frac { -b }{ a }\)
αβ + βγ + γα = – 7 = \(\frac { c }{ a }\)
αβγ = – 6 = \(\frac { -d }{ a }\)
Putting the values of   \(\frac { b }{ a }\), \(\frac { c }{ a }\),  and \(\frac { d }{ a }\)  in (1), we get
x3 – (0) x2 + (–7)x + (–6)
⇒ x3 – 7x + 6

Example 8:   If α and β are the zeroes of the polynomials  ax2 + bx + c then form the polynomial whose zeroes are    \(\frac { 1 }{ \alpha  } \quad and\quad \frac { 1 }{ \beta  } \)
Since α and β are the zeroes of ax2 + bx + c
So α + β = \(\frac { -b }{ a }\) ,     α β =  \(\frac { c }{ a }\)
Sum of the zeroes = \(\frac { 1 }{ \alpha  } +\frac { 1 }{ \beta  } =\frac { \alpha +\beta  }{ \alpha \beta  }  \)
\(=\frac{\frac{-b}{c}}{\frac{c}{a}}=\frac{-b}{c}\)
Product of the zeroes
\(=\frac{1}{\alpha }.\frac{1}{\beta }=\frac{1}{\frac{c}{a}}=\frac{a}{c}\)
But required polynomial is
x2 – (sum of zeroes) x + Product of zeroes
\(\Rightarrow {{\text{x}}^{2}}-\left( \frac{-b}{c} \right)\text{x}+\left( \frac{a}{c} \right)\)
\(\Rightarrow {{\text{x}}^{2}}+\frac{b}{c}\text{x}+\frac{a}{c}\)
\(\Rightarrow c\left( {{\text{x}}^{2}}+\frac{b}{c}\text{x}+\frac{a}{c} \right)\)
⇒ cx2 + bx + a

How Do You Determine The Degree Of A Polynomial

Degree Of A Polynomial

The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial.
For example :
In polynomial 5x2 – 8x7 + 3x:
(i) The power of term 5x2 = 2
(ii) The power of term –8x7 = 7
(iii) The power of 3x = 1
Since, the greatest power is 7, therefore degree of the polynomial 5x2 – 8x7 + 3x is 7
The degree of polynomial :
(i) 4y3 – 3y + 8 is 3
(ii) 7p + 2 is 1(p = p1)
(iii) 2m – 7m8 + m13 is 13 and so on.

Degree Of A Polynomial With Example Problems With Solutions

Example 1:    Find which of the following algebraic expression is a polynomial.
(i) 3x2 – 5x        (ii) \(\text{x + }\frac{1}{\text{x}}\)     (iii) √y– 8             (iv) z5 – ∛z + 8
Sol.
(i) 3x2 – 5x = 3x2 – 5x1
It is a polynomial.
(ii) \(\text{x + }\frac{1}{\text{x}}\) = x1 + x-1
It is not a polynomial.
(iii) √y– 8 = y1/2– 8
Since, the power of the first term (√y) is \(\frac{1}{2}\), which is not a whole number.
(iv) z5 – ∛z + 8 = z5 – z1/3 + 8
Since, the exponent of the second term is 1/3, which in not a whole number. Therefore, the given expression is not a polynomial.

Example 2:    Find the degree of the polynomial :
(i) 5x – 6x3 + 8x7 + 6x2    (ii) 2y12 + 3y10 – y15 + y + 3   (iii) x    (iv) 8
Sol.
(i)  Since the term with highest exponent (power) is 8x7 and its power is 7.
∴ The degree of given polynomial is 7.
(ii)  The highest power of the variable is 15
∴ degree = 15
(iii)  x = x1   ⇒   degree is 1.
(iv)  8 = 8x0   ⇒   degree = 0

Zeros Of A Polynomial Function

Zeros Of A Polynomial Function

If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x).

For Example:
(i) For polynomial p(x) = x – 2; p(2) = 2 – 2 = 0
∴ x = 2 or simply 2 is a zero of the polynomial
p(x) = x – 2.
(ii) For the polynomial g(u) = u2 – 5u + 6;
g(3) = (3)2 – 5 × 3 + 6 = 9 – 15 + 6 = 0
∴ 3 is a zero of the polynomial g(u)
= u2 – 5u + 6.
Also, g(2) = (2)2 – 5 × 2 + 6 = 4 – 10 + 6 = 0
∴ 2 is also a zero of the polynomial
g(u) = u2 – 5u + 6
(a) Every linear polynomial has one and only one zero.
(b) A given polynomial may have more than one zeroes.
(c) If the degree of a polynomial is n; the largest number of zeroes it can have is also n.
For Example:
If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a
polynomial is 8; largest number of zeroes it can have is 8.
(d) A zero of a polynomial need not be 0.
For Example: If f(x) = x2 – 4,
then f(2) = (2)2 – 4 = 4 – 4 = 0
Here, zero of the polynomial f(x) = x2 – 4 is 2 which itself is not 0.
(e) 0 may be a zero of a polynomial.
For Example: If f(x) = x2 – x,
then f(0) = 02 – 0 = 0
Here 0 is the zero of polynomial
f(x) = x2 – x.

Zeros Of A Polynomial Function With Examples

Example 1:    Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :
(i) p(x) = 3x + 1, x = \(-\frac{1}{3}\)
(ii) p(x) = (x + 1) (x – 2), x = – 1, 2
(iii) p(x) = x2, x = 0
(iv) p(x) = lx + m, x = \(-\frac{m}{l}\)
(v) p(x) = 2x + 1, x = \(\frac{1}{2}\)
Sol.
(i) p(x) = 3x + 1
\(\Rightarrow p\left( -\frac{1}{3} \right)=3\times -\frac{1}{3}+1=-1+1=0\)
∴ x = \(-\frac{1}{3}\)  is a zero of p(x) = 3x + 1.
(ii) p(x) = (x + 1) (x – 2)
⇒ p(–1) = (–1 + 1) (–1 – 2) = 0 × –3 = 0
and, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0
∴  x = –1 and x = 2 are zeroes of the given polynomial.
(iii) p(x) = x
⇒ p(0) = 02 = 0
∴  x = 0 is a zero of the given polynomial
(iv) p(x) = lx + m
\(\Rightarrow p\left( -\frac{m}{l} \right)=l\left( -\frac{m}{l} \right)+m\)
= – m + m = 0
∴  x = \(-\frac{m}{l}\)  is a zero of the given polynomial.
(v) p(x) = 2x + 1
\(\Rightarrow p\left( \frac{1}{2} \right)=2\times \frac{1}{2}+1\)
= 1 + 1 = 2 ≠ 0
∴ x = \(\frac{1}{2}\) is not a zero of the given polynomial.

Example 2:    Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = 2x + 5
(iii) p(x) = 3x – 2
Sol.
To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.
(i) For the zero of polynomial p(x) = x + 5
p(x) = 0      ⇒   x + 5 = 0     ⇒   x = –5
∴   x = –5 is a zero of the polynomial.
p(x) = x + 5.
(ii) p(x) = 0 ⇒ 2x + 5 = 0
⇒ 2x = –5 and x = \(-\frac{5}{2}\)
∴  \(-\frac{5}{2}\) is a zero of p(x) = 2x + 5.
(iii) p(x) = 0 ⇒ 3x – 2 = 0
⇒ 3x = 2 and x = \(\frac{2}{3}\).
∴  x = \(\frac{2}{3}\) is zero of p(x) = 3x – 2.

Types Of Polynomials

Types Of Polynomials

(i) Based on degree :
If degree of polynomial is

Examples
1.One  Linearx + 3, y – x + 2, √3x –3
2.TwoQuadratic2x2 –7, \(\frac{1}{3}{{\text{x}}^{2}}+{{\text{y}}^{2}}-2\text{xy}\), x2 +1+ 3y
3.ThreeCubicx3 + 3x2 –7x+8, 2x2+5x3+7,
4.Fourbi-quadraticx4 + y4 + 2x2y2, x4 + 3,…

(ii) Based on Terms :
If number of terms in polynomial is

 Examples
1.OneMonomial7x, 5x9, 3x16, xy, ……
2.TwoBinomial2 + 7y6, y3 + x14, 7 + 5x9,…
3.ThreeTrinomialx3 –2x + y, x31+y32+ z33,…..

Note:
(1) Degree of constant polynomials (Ex.5, 7, –3, 8/5, …) is zero.
(2) Degree of zero polynomial (zero = 0 = zero polynomial) is not defined.

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Algebraic Expression

Algebraic Expression

Any combination of literal numbers or variables and numbers (numerals) connected by +, –, × or ÷ signs is called an algebraic expression.
For example,
5, 6x, a + b × c, 4 × m + n, x – y ÷ z are algebraic expressions. The perimeter P of a triangle whose sides are a, b and c is given by P = a + b + c, area of square is x × x i.e., x2 are algebraic expressions.
A repeated product of a number with itself is written in exponential form.
2 × 2 × 2 × 2 = 24, etc
The same is true for literal numbers also. Thus, if x is a literal, then we have
x × x × x = x3 (Third power of x or x cube)
x × x × x × x = x4 and so on.
Also,           7 × x × x = 7x2
4 × x × x × y × y = 4x2 y2, etc.
x × x × x ……, n times = xn and read as n-th power of x. Here x is called base and n, exponent.

For example :

Algebraic ExpressionNo. of termsTerms
(i)   –32x1–32x
(ii)  2x + 3y22x and 3y
(iii) ax – 5y + cz3ax, –5y and cz
(iv) \(\frac{3}{x}+\frac{y}{7}-\frac{xy}{8}+9\)4 \(\frac{3}{x},\frac{y}{7},-\frac{xy}{8},9\text{ }\And \text{ }so\text{ }on\)

Factors:
When two or more numbers are multiplied together, the numbers themselves are called the factors of the product. The factors of 34 are 2 and 17. The factors of 18x2 are 2, 3 and x, factors of ln are l and n (8 is numerical factor of 8x and x is variable factor of 8x)

Coefficient:
Any factor of a product is the coefficient of the remaining factors. In the product of 8 × 5, the number 8 is coefficient of 5 and 5 is coefficient of 8. In the product 5yz, 5 is the (numerical) coefficient of yz, 5y is the coefficient of z and 5z is coefficient of y and yz is the variable coefficient of 5.
Note : If a letter has no coefficient written before it, the coefficient 1 is understood. Thus y means 1y and yz means 1yz, similarly, –y means –1(y).

Terms

An algebraic expression is a combination of numbers, literals and arithmetical operations. One or more sings (+ and –) separates an expression into several parts. Each part along with its sign is called a term.

Type of Algebraic Expression Definition Examples
1. MonomialA monomial is an expression having one term.6, –5xy, –6x2 etc.
2. BinomialA binomial is an algebraic expression having two terms.2x – 3y, x – y, 3x2 – 6x,
(x – y)2 + 3xy, 3x2 + 5 etc.
3. TrinomialA trinomial is an algebraic expression having three terms.2a – 3b – 5c, 5y2 – 3x + 9, a3 + b3 + c3 etc.
4.Tetranomial  or QuadrinomialA quadrinomial is an algebraic expression having four terms.a + b + c – 3, a3 + b3 + c3 + 3abc, etc.
5. PolynomialBinomial, trinomials and all algebraic expressions having more than three  terms are called polynomials.2a – 3b, x + y – 3yz + 4x2 – 6y2 etc.

Note : 
(i) The words ‘mono’, ‘bi’, ‘tri’ and ‘poly’ mean one, two, three and many.
(ii) A term of an algebraic expression having no literal factor is called a constant term, for example, in x2 + 9x – 8, the constant term is –8.

Example 1: Classify the following algebraic expressions into monomial, binomial, and trinomial:
a + b, 4x + 3y – 7z, y2, 1 + x + x2, z2 + 2
Solution: Monomial: y2
Binomials: a + b, z2 + 2
Trinomials: 4x + 3y – 7z, 1 + x + x2

Example 2: Find the value of a2 + x2, if a = 2 and x = 3.
Solution: a2 + x2 = a × a + x × x
= 2x+ 3x3
= 4 + 9 = 13

Example 3: Write an algebraic expression whose terms are
(a) 7, 4x, 3xy      (b) 4xy2, 3x2y, -9
Solution:
(a) The algebraic expression with terms
7, 4x, and 3xy = 7 + 4x + 3xy
(b) The algebraic expression with terms
4xy2, 3x2y, and – 9 = 4xy2 + 3x2y + (- 9)
= 4xy2 + 3x2y – 9

Like (Similar) and Unlike (Dissimilar ) Terms

These terms are defined as follows:

Like (Similar terms)Unlike (Dissimilar terms)
These are terms whose literal (variable) factors are the same.
For example,
(i) 5x2, – 6x2, + 3x2
(ii) 2(a + b), – 4(a + b), 6(a + b)
(iii) 6xy2, –8xy2, xy2
These are terms whose literal (variable) factors are not same.
For example,
(i) 2x and 5y
(ii) 6xy2 and 8x2y
(iii) (x + y), (x2 + y), 5(x2 + y2)

Finding the Value of an Algebraic Expression

An algebraic expression contains literal (variable) numbers. If we know the numerical values of these variables and substitute them in the given algebraic expression we get a numerical expression which can be simplified by the methods of arithmetic to get a number, called the value of the algebraic expression.

Eg: Evaluate the following expression if a = 2, b = – 1,  c = 1: a2 + b2 + c2 – ab – bc – ca.
Solution:
a2 + b2 + c2 – ab – bc – ca
= (2)2 + (–1)2 + (1)2 – 2 × (–1) – (–1) × 1 – 1 × 2
= 4 + 1 + 1 + 2 + 1 – 2
= 9 – 2 = 7
So, a2 + b2 + c2 – ab – bc – ca at a = 2, b = – 1,  c = 1 has value 7.

Addition and Subtraction of Algebraic Expressions

Like Terms :

  1. The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
  2. The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.

Addition and Subtraction of Algebraic Expressions Problems with Solutions

1. Add 2x2, 5x2 and x2.
Solution:
The three like terms are 2x2, 5x2 and x2
Adding these terms, we have 2x2 + 5x2 + x2
=  (2 + 5 + 1) x2      (Distributive property)
=   8x2

2. Add – xy, –5xy, –2xy.
Solution:
Adding the given three like terms, we have
= – xy + (– 5xy) + (– 2xy) = –xy –5xy – 2xy
= (–1 – 5 – 2) xy = – 8xy

3. Add –7x, –5x, 8x, 9x.
Solution:
Positive terms are 8x, 9x
Negative terms are – 7x, – 5x
Sum of positive terms = 8x + 9x = 17x
Sum of negative terms = – 7x + (–5x)
= – 7x – 5x = – 12x
Adding these two terms, we have
17x + (–12)x = 17x – 12x = (17 – 12)x = 5x

4. Add 4x + 3y – 5z, –7z + 5x – 8y and – y –3x + 2z.
Solution:
Column method:
Algebraic Expression 1
Re-write the expressions so that their like terms are in a column as
Horizontal Method :
Sum = (4x + 3y – 5z) + (–7z + 5x – 8y) + (–y –3x + 2z)
= 4x + 3y – 5z –7z + 5x – 8y – y –3x + 2z
= (4x + 5x –3x) + (3y – 8y – y) + ( – 5z – 7z +2z)
= (4 + 5 – 3) x + (3 – 8 – 1) y + (–5 –7 + 2)z
= 6x – 6y –10z

5. Subtract 10x2 from –8x2.
Solution:
Algebraic Expression 2

6. How much is t2 – 5t + 6 greater than t2 + 5t – 6 ?
Solution:
Algebraic Expression 3

Rule for Solving Algebraic Expressions

BODMAS represents the order of Performance of operations namely BODMAS represents the order of Performance of operations namely
B = Brackets;
O = Of;
D = Division;
M = Multiplication;
A = Addition;
S = Subtraction

Examples:

1. Simplify : 2x – [3y – {2x – (y – x)}].
Solution:
We have, 2x – [3y – {2x – (y – x)}]
We first remove the inner most bracket.
2x – [3y – {2x – y + x}]
Next inner most is the curly bracket.
2x – [3y – 2x + y – x]
Now we remove the square bracket.
2x – 3y + 2x – y + x
= (2x + 2x + x) – 3y – y
= (2x + 2x + x) – (3y + y)
= (2 + 2 + 1) x – (3 + 1) y = 5x – 4y

2. Simplify : 5a – [a2 – {2a (1 – a + 4a2) –3a (a2 – 5a – 3)}] – 8a.
Solution:
We first remove the inner most grouping symbol ( ), the { } and then [ ].
Thus we have,
5a – [a2 – {2a (1 – a + 4a2) –3a (a2 – 5a – 3)}] – 8a
= 5a –[a2 –{2a –2a2 + 8a3 – 3a3 + 15a2 + 9a}] – 8a
= 5a –[a2 – 2a + 2a2 – 8a3 + 3a3 – 15a2 – 9a] – 8a
= 5a – a2 + 2a – 2a2 + 8a3 – 3a3 + 15a2 + 9a – 8a
= 5a3 + 12a2 + 8a.