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**Revision Notes on Polynomials**

**Polynomial**

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.

**Example**

1. 4x^{3} + 3x^{2} + x +3 is a polynomial in variable x.

2. 4x^{2} + 3x^{-1} – 4 is not a polynomial as it has negative power.

3. 3x^{3/2 }+ 2x – 3 is not a polynomial.

- Polynomials are denoted by p (x), q (x), etc.
- In the above polynomials, 2x
^{2}, 3y, and 2 are the terms of the polynomial. - 2 and 3 are the coefficients of x
^{2}and y, respectively. - x and y are the variables.
- 2 is the constant term, which has no variable.

**Polynomials in One Variable**

If there is only one variable in the expression, then this is called the polynomial in one variable.

**Example**

- x
^{3}+ x – 4 is polynomial in variable x and is denoted by p(x). - r
^{2}+ 2 is polynomial in variable r and is denoted by p(r).

**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.

**Zeros of a 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**.

**Example: 1**

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.

**Example: 2**

Find p (1) for the polynomial p (t) = t^{2} – t + 1

p (1) = (1)^{2} – 1 + 1

= 1 – 1 + 1

= 1

**Remainder Theorem**

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

**Division of a Polynomial with a Monomial**

We can see that ‘x’ is common in the above polynomial, so we can write it as

Hence, 3x^{2 }+ x + 1 and x the factors of 3x^{3} + x^{2} + x.

**Steps of the Division of a Polynomial with a Non –Zero Polynomial**

Divide x^{2} – 3x -10 by 2 + x

**Step 1:** Write the dividend and divisor in descending order, i.e., in the standard form. x^{2} – 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x^{2}/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 x^{2} – 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.

**Example:**

What is the remainder if a^{4} + a^{3} – 2a^{2} + a + 1 is divided by a – 1.

**Solution:**

P(x) = a^{4} + a^{3} – 2a^{2} + 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.

**Factor Theorem**

The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

- (y – t) is a factor of p(y), if p(t) = 0, and
- P (t) = 0 if (y – t) is a factor of p (y).

**Example: 1**

Check whether g(x) = x – 3 is the factor of p(x) = x^{3 }– 4x^{2 }+ x + 6 using the factor theorem.

**Solution:**

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).

**Example: 2**

Find the value of k, if x – 1 is a factor of p(x) = kx^{2 }– √2x + 1

**Solution:**

As x -1 is the factor, p(1) = 0

**Factorization of Polynomials**

Factorization can be done by three methods

**1. By taking out the common factor**

If we have to factorise x^{2} –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x^{2} – 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**

x^{2} + bx + c = x^{2} + (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’.

**Example: 1**

Factorize 6x^{2} + 17x + 5 by splitting the middle term.

**Solution:**

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.

6x^{2} + 17x + 5 =6 x^{2} + (2 + 15) x + 5

= 6 x^{2} + 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Algebraic Identities |

1. (x + y)^{2 }= x^{2 }+ 2xy + y^{2} |

2. (x – y)^{2} = x^{2 }– 2xy + y^{2} |

3. (x + y) (x – y) = x^{2} – y^{2} |

4. (x + a) (x + b) = x^{2} + (a + b)x + ab |

5. (x + y + z)^{2 }= x^{2 }+ y^{2 }+ z^{2 }+ 2xy + 2yz + 2zx |

6. (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y) = x^{3}+ y^{3} + 3x^{2}y + 3xy^{2} |

7. (x – y)^{3} = x^{3}– y^{3} – 3xy(x – y) = x^{3 }– y^{3} – 3x^{2}y + 3xy^{2} |

8. x^{3 }+ y^{3} = (x + y)(x^{2} – xy + y^{2}) |

9. x^{3} – y^{3 }= (x – y)(x^{2} + xy + y^{2}) |

10. x^{3 }+ y^{3} + z^{3 }– 3xyz = (x + y + z)(x^{2} + y^{2 }+ z^{2} – xy – yz – zx) x^{3} + y^{3} + z^{3} = 3xyz if x + y + z = 0 |

**Example: 2**

Factorize 8x^{3} + 27y^{3} + 36x^{2}y + 54xy^{2}

**Solution:**

The given expression can be written as

= (2x)^{3 }+ (3y)^{3} + 3(4x^{2}) (3y) + 3(2x) (9y^{2})

= (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.

**Example: 3**

Factorize 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz.

**Solution:**

4x^{2} + y^{2} + z^{2} – 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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