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A polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral powers.
1. 4x3 + 3x2 + x +3 is a polynomial in variable x.
2. 4x2 + 3x-1 – 4 is not a polynomial as it has negative power.
3. 3x3/2 + 2x – 3 is not a polynomial, since it power is not an integer.
.
If there is only one variable in the expression, then this is called the polynomial in one variable.
Types of polynomials on the basis of the number of terms
Types of polynomials on the basis of the number of degrees
The highest value of the power of the variable in the polynomial is the degree of the polynomial.
If p(x) is a polynomial, then the number ‘a’ will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by equating it to zero.
The given polynomial is p(x) = x – 4
To find the zero of the polynomial, we will equate it to zero.
x – 4 = 0
x = 4
p(4) = x – 4 = 4 – 4 = 0
This shows that if we place 4 in place of x, we get the value of the polynomial as zero. So 4 is the zero of this polynomial. And also, we are getting the value 4 by equating the polynomial with 0.
So 4 is the zero of the polynomial or the root of the polynomial.
The root of the polynomial is basically the x-intercept of the polynomial.
If the polynomial has one root, it will intersect the x-axis at one point only, and if it has two roots, it will intersect at two points, and so on.
Find p (1) for the polynomial p (t) = t2 – t + 1
p (1) = (1)2 – 1 + 1
= 1 – 1 + 1
= 1
We know the property of division which follows in the basic division, i.e.
Dividend = (Divisor × Quotient) + Remainder
This follows the division of polynomials.
If p(x) and g(x) are two polynomials in which the degrees of p(x) ≥ degree of g(x) and g(x) ≠ 0 are given, then we can get the q(x) and r(x) so that:
P(x) = g(x) q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
It says that p(x) divided by g(x), gives q(x) as a quotient and r(x) as a remainder.
Let’s understand it with an example
We can see that ‘x’ is common in the above polynomial, so we can write it as
Hence, 3x2 + x + 1 and x the factors of 3x3 + x2 + x.
Divide x2 – 3x -10 by 2 + x
Step 1: Write the dividend and divisor in descending order, i.e., in the standard form. x2 – 3x -10 and x + 2
Divide the first term of the dividend with the first term of the divisor.
x2/x = x this will be the first term of the quotient.
Step 2: Now multiply the divisor by this term of the quotient and subtract it from the dividend.
Step 3: Now the remainder is our new dividend, so we will repeat the process again by dividing the dividend by the divisor.
Step 4: – (5x/x) = – 5
Step 5:
The remainder is zero.
Hence x2 – 3x – 10 = (x + 2)(x – 5) + 0
Dividend = (Divisor × Quotient) + Remainder
The Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).
As we know,
P(x) = g(x) q(x) + r(x)
If p(x) is divided by (x-t) then
If x = t
P (t) = (t – t). q (t) + r = 0
To find the remainder or to check the multiple of the polynomial, we can use the remainder theorem.
What is the remainder if a4 + a3 – 2a2 + a + 1 is divided by a – 1.
P(x) = a4 + a3 – 2a2 + a + 1
To find the zero of (a – 1), we need to equate it to zero.
a -1 = 0
a = 1
p (1) = (1)4 + (1)3 – 2(1)2 + (1) + 1
= 1 + 1 – 2 + 1 + 1
= 2
So by using the remainder theorem, we can easily find the remainder after the division of the polynomial.
The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then
Check whether g(x) = x – 3 is the factor of p(x) = x3 – 4x2 + x + 6 using the factor theorem.
According to the factor theorem, if x – 3 is the factor of p(x), then p(3) = 0, as the root of x – 3 is 3.
P (3) = (3)3 – 4(3)2 + (3) + 6
= 27 – 36 + 3 + 6 = 0
Hence, g (x) is the factor of p (x).
Find the value of k, if x – 1 is a factor of p(x) = kx2 – √2x + 1
As x -1 is the factor, p(1) = 0
Factorization can be done by three methods
1. By taking out the common factor
If we have to factorise x2 –x then we can do it by taking x common.
x(x – 1) so that x and x-1 are the factors of x2 – x.
2. By grouping
ab + bc + ax + cx = (ab + bc) + (ax + cx)
= b(a + c) + x(a + c)
= (a + c)(b + x)
3. By splitting the middle term
x2 + bx + c = x2 + (p + q) + pq
= (x + p)(x + q)
This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.
Factorize 6x2 + 17x + 5 by splitting the middle term.
If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.
Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.
6x2 + 17x + 5 =6 x2 + (2 + 15) x + 5
= 6 x2 + 2x + 15x + 5
= 2 x (3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
| Algebraic Identities |
| 1. (x + y)2 = x2 + 2xy + y2 |
| 2. (x – y)2 = x2 – 2xy + y2 |
| 3. (x + y) (x – y) = x2 – y2 |
| 4. (x + a) (x + b) = x2 + (a + b)x + ab |
| 5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx |
| 6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2 |
| 7. (x – y)3 = x3– y3 – 3xy(x – y) = x3 – y3 – 3x2y + 3xy2 |
| 8. x3 + y3 = (x + y)(x2 – xy + y2) |
| 9. x3 – y3 = (x – y)(x2 + xy + y2) |
| 10. x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) x3 + y3 + z3 = 3xyz if x + y + z = 0 |
Factorize 8x3 + 27y3 + 36x2y + 54xy2
The given expression can be written as
= (2x)3 + (3y)3 + 3(4x2) (3y) + 3(2x) (9y2)
= (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2
= (2x + 3y)3 (Using Identity VI)
= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.
Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)
= [2x + (- y) + z]2 (Using Identity V)
= (2x – y + z)2 = (2x – y + z) (2x – y + z)
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A polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral powers.
1. 4x3 + 3x2 + x +3 is a polynomial in variable x.
2. 4x2 + 3x-1 – 4 is not a polynomial as it has negative power.
3. 3x3/2 + 2x – 3 is not a polynomial.
If there is only one variable in the expression, then this is called the polynomial in one variable.
Types of polynomials on the basis of the number of terms
Types of polynomials on the basis of the number of degrees
The highest value of the power of the variable in the polynomial is the degree of the polynomial.
If p(x) is a polynomial, then the number ‘a’ will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by equating it to zero.
The given polynomial is p(x) = x – 4
To find the zero of the polynomial, we will equate it to zero.
x – 4 = 0
x = 4
p(4) = x – 4 = 4 – 4 = 0
This shows that if we place 4 in place of x, we get the value of the polynomial as zero. So 4 is the zero of this polynomial. And also, we are getting the value 4 by equating the polynomial with 0.
So 4 is the zero of the polynomial or the root of the polynomial.
The root of the polynomial is basically the x-intercept of the polynomial.
If the polynomial has one root, it will intersect the x-axis at one point only, and if it has two roots, it will intersect at two points, and so on.
Find p (1) for the polynomial p (t) = t2 – t + 1
p (1) = (1)2 – 1 + 1
= 1 – 1 + 1
= 1
We know the property of division which follows in the basic division, i.e.
Dividend = (Divisor × Quotient) + Remainder
This follows the division of polynomials.
If p(x) and g(x) are two polynomials in which the degrees of p(x) ≥ degree of g(x) and g(x) ≠ 0 are given, then we can get the q(x) and r(x) so that:
P(x) = g(x) q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
It says that p(x) divided by g(x), gives q(x) as a quotient and r(x) as a remainder.
Let’s understand it with an example
We can see that ‘x’ is common in the above polynomial, so we can write it as
Hence, 3x2 + x + 1 and x the factors of 3x3 + x2 + x.
Divide x2 – 3x -10 by 2 + x
Step 1: Write the dividend and divisor in descending order, i.e., in the standard form. x2 – 3x -10 and x + 2
Divide the first term of the dividend with the first term of the divisor.
x2/x = x this will be the first term of the quotient.
Step 2: Now multiply the divisor by this term of the quotient and subtract it from the dividend.
Step 3: Now the remainder is our new dividend, so we will repeat the process again by dividing the dividend by the divisor.
Step 4: – (5x/x) = – 5
Step 5:
The remainder is zero.
Hence x2 – 3x – 10 = (x + 2)(x – 5) + 0
Dividend = (Divisor × Quotient) + Remainder
The Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).
As we know,
P(x) = g(x) q(x) + r(x)
If p(x) is divided by (x-t) then
If x = t
P (t) = (t – t). q (t) + r = 0
To find the remainder or to check the multiple of the polynomial, we can use the remainder theorem.
What is the remainder if a4 + a3 – 2a2 + a + 1 is divided by a – 1.
P(x) = a4 + a3 – 2a2 + a + 1
To find the zero of (a – 1), we need to equate it to zero.
a -1 = 0
a = 1
p (1) = (1)4 + (1)3 – 2(1)2 + (1) + 1
= 1 + 1 – 2 + 1 + 1
= 2
So by using the remainder theorem, we can easily find the remainder after the division of the polynomial.
The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then
Check whether g(x) = x – 3 is the factor of p(x) = x3 – 4x2 + x + 6 using the factor theorem.
According to the factor theorem, if x – 3 is the factor of p(x), then p(3) = 0, as the root of x – 3 is 3.
P (3) = (3)3 – 4(3)2 + (3) + 6
= 27 – 36 + 3 + 6 = 0
Hence, g (x) is the factor of p (x).
Find the value of k, if x – 1 is a factor of p(x) = kx2 – √2x + 1
As x -1 is the factor, p(1) = 0
Factorization can be done by three methods
1. By taking out the common factor
If we have to factorise x2 –x then we can do it by taking x common.
x(x – 1) so that x and x-1 are the factors of x2 – x.
2. By grouping
ab + bc + ax + cx = (ab + bc) + (ax + cx)
= b(a + c) + x(a + c)
= (a + c)(b + x)
3. By splitting the middle term
x2 + bx + c = x2 + (p + q) + pq
= (x + p)(x + q)
This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.
Factorize 6x2 + 17x + 5 by splitting the middle term.
If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.
Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.
6x2 + 17x + 5 =6 x2 + (2 + 15) x + 5
= 6 x2 + 2x + 15x + 5
= 2 x (3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
| Algebraic Identities |
| 1. (x + y)2 = x2 + 2xy + y2 |
| 2. (x – y)2 = x2 – 2xy + y2 |
| 3. (x + y) (x – y) = x2 – y2 |
| 4. (x + a) (x + b) = x2 + (a + b)x + ab |
| 5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx |
| 6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2 |
| 7. (x – y)3 = x3– y3 – 3xy(x – y) = x3 – y3 – 3x2y + 3xy2 |
| 8. x3 + y3 = (x + y)(x2 – xy + y2) |
| 9. x3 – y3 = (x – y)(x2 + xy + y2) |
| 10. x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) x3 + y3 + z3 = 3xyz if x + y + z = 0 |
Factorize 8x3 + 27y3 + 36x2y + 54xy2
The given expression can be written as
= (2x)3 + (3y)3 + 3(4x2) (3y) + 3(2x) (9y2)
= (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2
= (2x + 3y)3 (Using Identity VI)
= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.
Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)
= [2x + (- y) + z]2 (Using Identity V)
= (2x – y + z)2 = (2x – y + z) (2x – y + z)
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Introduction :
Polynomial :
Any expression of the form a0xn+a1xn-1+a2xn-2+….an is called a polynomial of degree n in variable x ; a0≠0, where n is a non-negative integer and a0, a1, a2, ….., and are real numbers, called the coefficients of the terms of the polynomial. (Scroll down to continue …)
Audio, Visual & Digital Content
A polynomial in x can be denoted by the symbols p(x), q(x), f(x), g(x), etc.
Degree Of Polynomial: The highest power of x in p(x) is called the degree of the polynomial p(x).
Linear Polynomial : A polynomial of degree one is called a linear polynomial.
Quadratic Polynomial :
A polynomial of degree two is called a Quadratic Polynomial.
Generally, any quadratic polynomial in x is of the form ax2+bx+c, a ≠ 0 and a, b, c are real numbers.
Cubic Polynomial :
A polynomial of degree three is called a Cubic Polynomial.
Generally, any cubic polynomial in x is of the form ax3+bx2+cx+d, a≠0 and a, b, c, d are real numbers.
Value of a Polynomial :
If we replace x by ‘ -2’ in the polynomial p(x) = 3x3-2x2+x-1
we have p(-2) =3(-2)3-2(-2)2+(-2)-1
= -24-8-2-1 =-35
Thus, on replacing x by ‘ -2 ‘ in the polynomial p(x), we get -35, which is called the value of the polynomial.
Hence, if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of the polynomial p(x) at x=k, and generally, denoted by p(k).
Zeros of a Polynomial :
A real constant, k is said to be a zero of a polynomial p(x) in x, if p(k)=0
For example, the polynomial p(x) = x2+x-12 gives p(3)=32+3-12=0 and p(-4)=(-4)2+(-4)-12=0.
Thus, 3 and -4 are two zeroes of the polynomial p(x).
A linear polynomial (degree one) has one and only one zero, given by;
Zero of the linear polynomial = -(constant term )coefficient of x
Geometrical Representation of the Zeroes of a Polynomial :
Let us consider a linear polynomial p(x)=3x-6.
We know that, graph of a linear polynomial is a straight line.
Therefore, graph of p(x)=3x-6 is a straight line passing through the points (1,-3),(3,3),(2,0).
Table for p(x)=3x – 6
From the graph of p(x)=3x-6, we observe that it intersects the x-axis at the point (2,0).
Zero of the polynomial [p(x)=3x-6] = -(-6)3 = 63 = 2.
Thus, we conclude that the zero of the polynomial p(x) = 3x – 6 is the x-coordinate of the point where the graph of p(x) = 3x – 6 intersects the x-axis.
Similarly, the zeroes of a quadratic polynomial, p(x) = ax2+bx+c, a≠0, are the x-coordinates of the points where the graph (parabola) of p(x)=ax2+bx+c, a≠0, intersects the x-axis.
Graph of p(x) = ax2+bx+c, a≠0 intersects the x-axis at the most in two points and hence the quadratic polynomial can have at most two distinct real zeros.
A cubic polynomial can have at most three distinct real zeros.
Relation between Zeroes and Coefficients of a Polynomial :
Let the quadratic polynomial be p(x) = ax2+bx+c, a≠0 and having zeroes as α and β, then
Sum of the zeroes = α + β
= -(coefficient of x) /(coefficient of x2) = -b/a
Product of the zeroes = αβ
Let the cubic polynomial be p(x) = ax3+bx2+cx+d, a≠0 and having zeroes as α , β and γ, then Sum of the zeroes = α + β + γ
α + β + γ = -(coefficient of x2 )/(coefficient of x3)= -b/a
αβ = (constant term) /(coefficient of x2) = c/a
Sum of the products of zeroes taken two at a time αβ+βγ+γα
αβ+βγ+γα = (coefficient of x) /(coefficient of x3)= c/a
and
Product of the zeroes = αβγ
αβγ = (constant term) /(coefficient of x3)= -d/a
Division Algorithm for Polynomials :
For any two polynomials p(x) and g(x) ; g(x) ≠0, we can find two polynomials q(x) and r(x), such that p(x)=g(x) × q(x)+r(x).
Where r(x)=0 or degree of r(x) is less than the degree of g(x). Here, q(x) is called quotient, r(x) is called remainder, p(x) is called dividend and g(x) is called divisor. This result is known as a division algorithm for polynomials.
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Introduction :
Polynomial :
Any expression of the form a0xn+a1xn-1+a2xn-2+….an is called a polynomial of degree n in variable x ; a0≠0, where n is a non-negative integer and a0, a1, a2, ….., and are real numbers, called the coefficients of the terms of the polynomial. (Scroll down to continue …)
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A polynomial in x can be denoted by the symbols p(x), q(x), f(x), g(x), etc.
Degree Of Polynomial: The highest power of x in p(x) is called the degree of the polynomial p(x).
Linear Polynomial : A polynomial of degree one is called a linear polynomial.
Quadratic Polynomial :
A polynomial of degree two is called a Quadratic Polynomial.
Generally, any quadratic polynomial in x is of the form ax2+bx+c, a ≠ 0 and a, b, c are real numbers.
Cubic Polynomial :
A polynomial of degree three is called a Cubic Polynomial.
Generally, any cubic polynomial in x is of the form ax3+bx2+cx+d, a≠0 and a, b, c, d are real numbers.
Value of a Polynomial :
If we replace x by ‘ -2’ in the polynomial p(x) = 3x3-2x2+x-1
we have p(-2) =3(-2)3-2(-2)2+(-2)-1
= -24-8-2-1 =-35
Thus, on replacing x by ‘ -2 ‘ in the polynomial p(x), we get -35, which is called the value of the polynomial.
Hence, if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of the polynomial p(x) at x=k, and generally, denoted by p(k).
Zeros of a Polynomial :
A real constant, k is said to be a zero of a polynomial p(x) in x, if p(k)=0
For example, the polynomial p(x) = x2+x-12 gives p(3)=32+3-12=0 and p(-4)=(-4)2+(-4)-12=0.
Thus, 3 and -4 are two zeroes of the polynomial p(x).
A linear polynomial (degree one) has one and only one zero, given by;
Zero of the linear polynomial = -(constant term )coefficient of x
Geometrical Representation of the Zeroes of a Polynomial :
Let us consider a linear polynomial p(x)=3x-6.
We know that, graph of a linear polynomial is a straight line.
Therefore, graph of p(x)=3x-6 is a straight line passing through the points (1,-3),(3,3),(2,0).
Table for p(x)=3x – 6
From the graph of p(x)=3x-6, we observe that it intersects the x-axis at the point (2,0).
Zero of the polynomial [p(x)=3x-6] = -(-6)3 = 63 = 2.
Thus, we conclude that the zero of the polynomial p(x) = 3x – 6 is the x-coordinate of the point where the graph of p(x) = 3x – 6 intersects the x-axis.
Similarly, the zeroes of a quadratic polynomial, p(x) = ax2+bx+c, a≠0, are the x-coordinates of the points where the graph (parabola) of p(x)=ax2+bx+c, a≠0, intersects the x-axis.
Graph of p(x) = ax2+bx+c, a≠0 intersects the x-axis at the most in two points and hence the quadratic polynomial can have at most two distinct real zeros.
A cubic polynomial can have at most three distinct real zeros.
Relation between Zeroes and Coefficients of a Polynomial :
Let the quadratic polynomial be p(x) = ax2+bx+c, a≠0 and having zeroes as α and β, then
Sum of the zeroes = α + β
= -(coefficient of x) /(coefficient of x2) = -b/a
Product of the zeroes = αβ
Let the cubic polynomial be p(x) = ax3+bx2+cx+d, a≠0 and having zeroes as α , β and γ, then Sum of the zeroes = α + β + γ
α + β + γ = -(coefficient of x2 )/(coefficient of x3)= -b/a
αβ = (constant term) /(coefficient of x2) = c/a
Sum of the products of zeroes taken two at a time αβ+βγ+γα
αβ+βγ+γα = (coefficient of x) /(coefficient of x3)= c/a
and
Product of the zeroes = αβγ
αβγ = (constant term) /(coefficient of x3)= -d/a
Division Algorithm for Polynomials :
For any two polynomials p(x) and g(x) ; g(x) ≠0, we can find two polynomials q(x) and r(x), such that p(x)=g(x) × q(x)+r(x).
Where r(x)=0 or degree of r(x) is less than the degree of g(x). Here, q(x) is called quotient, r(x) is called remainder, p(x) is called dividend and g(x) is called divisor. This result is known as a division algorithm for polynomials.
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Audio, Visual & Digital Content
A polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral powers.
1. 4x3 + 3x2 + x +3 is a polynomial in variable x.
2. 4x2 + 3x-1 – 4 is not a polynomial as it has negative power.
3. 3x3/2 + 2x – 3 is not a polynomial.
If there is only one variable in the expression, then this is called the polynomial in one variable.
Types of polynomials on the basis of the number of terms
Types of polynomials on the basis of the number of degrees
The highest value of the power of the variable in the polynomial is the degree of the polynomial.
If p(x) is a polynomial, then the number ‘a’ will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by equating it to zero.
The given polynomial is p(x) = x – 4
To find the zero of the polynomial, we will equate it to zero.
x – 4 = 0
x = 4
p(4) = x – 4 = 4 – 4 = 0
This shows that if we place 4 in place of x, we get the value of the polynomial as zero. So 4 is the zero of this polynomial. And also, we are getting the value 4 by equating the polynomial with 0.
So 4 is the zero of the polynomial or the root of the polynomial.
The root of the polynomial is basically the x-intercept of the polynomial.
If the polynomial has one root, it will intersect the x-axis at one point only, and if it has two roots, it will intersect at two points, and so on.
Find p (1) for the polynomial p (t) = t2 – t + 1
p (1) = (1)2 – 1 + 1
= 1 – 1 + 1
= 1
We know the property of division which follows in the basic division, i.e.
Dividend = (Divisor × Quotient) + Remainder
This follows the division of polynomials.
If p(x) and g(x) are two polynomials in which the degrees of p(x) ≥ degree of g(x) and g(x) ≠ 0 are given, then we can get the q(x) and r(x) so that:
P(x) = g(x) q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
It says that p(x) divided by g(x), gives q(x) as a quotient and r(x) as a remainder.
Let’s understand it with an example
We can see that ‘x’ is common in the above polynomial, so we can write it as
Hence, 3x2 + x + 1 and x the factors of 3x3 + x2 + x.
Divide x2 – 3x -10 by 2 + x
Step 1: Write the dividend and divisor in descending order, i.e., in the standard form. x2 – 3x -10 and x + 2
Divide the first term of the dividend with the first term of the divisor.
x2/x = x this will be the first term of the quotient.
Step 2: Now multiply the divisor by this term of the quotient and subtract it from the dividend.
Step 3: Now the remainder is our new dividend, so we will repeat the process again by dividing the dividend by the divisor.
Step 4: – (5x/x) = – 5
Step 5:
The remainder is zero.
Hence x2 – 3x – 10 = (x + 2)(x – 5) + 0
Dividend = (Divisor × Quotient) + Remainder
The Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).
As we know,
P(x) = g(x) q(x) + r(x)
If p(x) is divided by (x-t) then
If x = t
P (t) = (t – t). q (t) + r = 0
To find the remainder or to check the multiple of the polynomial, we can use the remainder theorem.
What is the remainder if a4 + a3 – 2a2 + a + 1 is divided by a – 1.
P(x) = a4 + a3 – 2a2 + a + 1
To find the zero of (a – 1), we need to equate it to zero.
a -1 = 0
a = 1
p (1) = (1)4 + (1)3 – 2(1)2 + (1) + 1
= 1 + 1 – 2 + 1 + 1
= 2
So by using the remainder theorem, we can easily find the remainder after the division of the polynomial.
The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then
Check whether g(x) = x – 3 is the factor of p(x) = x3 – 4x2 + x + 6 using the factor theorem.
According to the factor theorem, if x – 3 is the factor of p(x), then p(3) = 0, as the root of x – 3 is 3.
P (3) = (3)3 – 4(3)2 + (3) + 6
= 27 – 36 + 3 + 6 = 0
Hence, g (x) is the factor of p (x).
Find the value of k, if x – 1 is a factor of p(x) = kx2 – √2x + 1
As x -1 is the factor, p(1) = 0
Factorization can be done by three methods
1. By taking out the common factor
If we have to factorise x2 –x then we can do it by taking x common.
x(x – 1) so that x and x-1 are the factors of x2 – x.
2. By grouping
ab + bc + ax + cx = (ab + bc) + (ax + cx)
= b(a + c) + x(a + c)
= (a + c)(b + x)
3. By splitting the middle term
x2 + bx + c = x2 + (p + q) + pq
= (x + p)(x + q)
This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.
Factorize 6x2 + 17x + 5 by splitting the middle term.
If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.
Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.
6x2 + 17x + 5 =6 x2 + (2 + 15) x + 5
= 6 x2 + 2x + 15x + 5
= 2 x (3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
| Algebraic Identities |
| 1. (x + y)2 = x2 + 2xy + y2 |
| 2. (x – y)2 = x2 – 2xy + y2 |
| 3. (x + y) (x – y) = x2 – y2 |
| 4. (x + a) (x + b) = x2 + (a + b)x + ab |
| 5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx |
| 6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2 |
| 7. (x – y)3 = x3– y3 – 3xy(x – y) = x3 – y3 – 3x2y + 3xy2 |
| 8. x3 + y3 = (x + y)(x2 – xy + y2) |
| 9. x3 – y3 = (x – y)(x2 + xy + y2) |
| 10. x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) x3 + y3 + z3 = 3xyz if x + y + z = 0 |
Factorize 8x3 + 27y3 + 36x2y + 54xy2
The given expression can be written as
= (2x)3 + (3y)3 + 3(4x2) (3y) + 3(2x) (9y2)
= (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2
= (2x + 3y)3 (Using Identity VI)
= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.
Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)
= [2x + (- y) + z]2 (Using Identity V)
= (2x – y + z)2 = (2x – y + z) (2x – y + z)
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