Polynomial: Definition, Value, Terms, Distribution and Example Problems

August 19, 2023
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Polynomial: Definition, Value, Conditions, Division and Example Problems – What is meant by Polynomial? On this occasion About the knowledge.co.id will discuss about Polynomials and the things that surround them. Let's look at the article below to understand it better.

Polynomial: Definition, Value, Terms, Distribution and Example Problems

Polynomials or commonly referred to as polynomials are a form of terms with many values composed of variable variables and constants. The operations used are only addition, subtraction, multiplication and powers of non-negative integers.

The general form of this Polynomial, namely:

Polynomial General Form: a_n xⁿ + a_n-1 x^n-1 +... + a₁ x + a

Information:

with a_n, a_n-1, …., a₁, a₀€ R coefficient or constant

Polynom a_n ≠ 0, and n are positive integers.

The highest power of x is the degree of the polynomial. While terms that do not contain a variable (a) are referred to as fixed (constant) terms.

A polynomial can look like the following:
25x² +19x – 06

Another example of the polynomial form is:

instagram viewer

3x
x – 2
-6y² – (½)x
3xyz + 3xy²z – 0.1xz – 200y + 0.5
512v⁵+99w⁵
5 (Constants are coefficients whose variable has a power of 0, so a number is a polynomial.)

A polynomial can have:

A variable (is a mutable value, like x, y, z in an equation; may have more than 1 variable)
Coefficients (are constants accompanying variables)
Constant (a fixed value that doesn't change)
The exponent or power is the power of the variable; can also be referred to as degrees of a polynomial.

Polynomial Terms

There are also several conditions so that an equation can be called a 'polynomial', including the following:

Variables cannot have fractional or negative exponents.
Variables cannot be included in a trigonometry equation.

Polynomial and Non-Polynomial

Here are some forms that are not included in the polynomial form, including the following:

3xy^-2, because the rank is negative. Exponents or powers can only be {0,1,2…}.
2/(x+2), because dividing by the variable is not allowed (the power of the denominator is negative).
1/x ,for the same reason ^.
√x, because the root is a power of a fraction, which is not allowed.
x cos x, because there is a variable x in trigonometric functions

Here are the things that are allowed or included in the polynomial form, pay close attention:

x/2 is allowed, because it's okay to divide by a constant.
√x² yes, because after explaining the result there is no exponential fraction.
√2 may be because the root is a constant, not a variable.
½ x⁵ – (cos∏)x³– (tan 60°)x – 1 is possible because trigonometric functions are constants, and there are no variables in them

Polynomial Value

We can find the value of the polynomial f (x) for x=k or f (k) using the substitution method or using the Horner scheme. Here are the details:

Substitution way:
By substituting x = k into the polynomial, it will become:

f(x) = a_n kⁿ + a_n-1 k^n-1 +... + a₁ k + a

How to horner scheme:
As an example:
(f(k) = x³ +bx² +cx +d so: f(k) = ak³ + bk² + ck + d
xa³ +bx² + cx + d = (ak² + bk + c) k+d
= ((ak + b) k + c) k+d

Polynomial Division

In general, division within a polynomial can be written as follows:

Formula: f(x) = g(x) h(x) + s(x)

Information:

f (x) is the polynomial that is divisible.
g (x) is the multiplier term.
h (x) is the polynomial term of the quotient.
s (x) is the remainder term.

Before we understand the polynomial division method, we must first know about the remainder theorem namely

Let F(x) be a polynomial of degree n,

If F(x) is divided (x-k) then the result is F(k)

If F(x) is divided (ax-b) then the result is F(b/a)

If F(x) is divided by (x-a)(x-b) then the result is:

Ordinary Distribution Method

An example is if 2x³ – 3x² + x + 5 divided by 2x² – x – 1

then the quotient and the remainder is the quotient = x-1 and the remainder = x+4

Horner's Division Method

We can divide polynomials f (x) by (x-k) using the Horner method.

We can use this method for degree 1 divisors or divisors that can be factored into degree 1 divisors.

The method is as follows:

Just write down the coefficient → it has to be coherent or sequential starting from the x coefficientⁿ, x^n–1, … to constants (if there is a non-existent variable, then the coefficient is written 0)

For example: for 4x³ – 1, the coefficients are 4, 0, 0, and -1 (for x³, x², x, and constants)

If the coefficient of the highest degree P(x) ≠ 1, then we must divide the quotient again by the coefficient of the highest degree P(x).
If we can factor the divisor, then:
- If the divisor can be factored to P₁ as well as P₂, then S(x) = P₁.S₂ + S₁
- If the divisor can be factored to P₁, P₂, P₃, then S(x) = P₁.P₂.S₃ +P₁.S₂ + S₁
- If the divisor can be factored into P₁, P₂, P₃, P₄, then S(x) = P₁.P₂.P₃.S₄ +P₁.P₂.S₃ +P₁.S₂ + S₁
- and so on.

Indefinite Coefficient Method

Basically, this method is done by substituting F(x) of degree m and P(x) of degree n into the general form of polynomial division, then filling H(x) and S(x) with

H(x) is a polynomial of degree k, where k = m – n

S(x) is a polynomial of degrees n-k

Examples of Polynomial Problems

question 1.

Is known

F(x) = 2x³ – 3x² + x + 5

P(x) = 2x² – x – 1

Determine the quotient and remainder

Answer :

F(x) = 2x³ – 3x² + x + 5

P(x) = 2x² – x – 1 = (2x + 1)(x – 1)

So p1: (2x + 1) = 0 -> x = -1/2 and p2: (x – 1) = 0 -> x = 1

Then the horner steps are shown in the following figure

So, the results are obtained and the remainder is as follows

H(x) = x-1

S(x) = P₁×S₂ + S₁ = x + 4

Problem 2.

Tribe of many x⁴ – 3x³ – 5x² + x – 6 divided by x² – x -2 the remainder equals …

a. 16x + 8
b. 16x – 8
c. -8x+16
d. -8x – 16
e. -8x – 24

Answer:

It is known that the divisor is: x² – x -2, so:
x² – x -2= 0
(x – 2) (x + 1) = 0
x = 2 and x = -1

Remember the formula: P(x) = H(x) + (px + q), so the remainder (px + q), then:

x = 2

f(2) = 2p + q
24 – 3(2)3 – 5(2)2 + 2 – 6 = 2p + q
16 – 24 – 20 + 2 – 6 = 2p + q
-32 = 2p + q … (i)

x = -1

f(-1) = -p + q
(-1) – 3(-1)3 – 5(-1)2 + (-1) – 6 = -p + q
1 + 4 – 5 – 1 – 6 = -p + q
-8 = -p + q …(ii)

Eliminate equations (i) and (ii), to become:

-32 =2p +q
-8 =-p +q
-24 =3p
p = -8

If we substitute p = –p + q = -8
-(-8) + q = -8
q = -16

So the remainder is = p + q = -8x – 16

Answer: D

Problem 3.

It is known that F(x) = 2x³ – 3x² + x + 5 ,P(x) = 2x² – x – 1

Determine the quotient and remainder using the indeterminate method

Discussion of questions:

m = 3, n = 2, k = 1

H(x) is degree 1, let's say H(x) = ax+b

S(x) is of degree 2-1=1 eg S(x) = px+q

Substitute F(x), P(x), H(x), S(x) into the equation

F(x) = P(x). H(x) + S(x), then obtained

2x³ – 3x² + x + 5 = (2x² – x – 1)(ax+b) + px+q

2x³ – 3x² + x + 5 = 2ax³+ 2bx²– ax²– bx – ax – b + px + q

(2)x³ +(– 3)x² + (1)x + (5) = (2a)x³+ (2b– a) x²+ (– b – a + p) x + (– b + q)

Then equate the coefficients of the left and right sides to be

2a = 2

a = 1

2b – a = -3

2b – 1 = -3

2b = -2

b = -1

– b – a + p = 1

1 – 1 + p = 1

p = 1

– b + q = 5

1 + q = 5

q = 4

So,

H(x) = ax + b = x – 1

S(x) = px + q = x + 4

Problem 4.

One of the factors of (2x³ -5x² – px =3) is (x + 1). Another factor of the multitude is…

a. (x – 2) and (x – 3)
b. (x + 2) and (2x – 1)
c. (x + 3) and (x + 2)
d. (2x + 1) and (x – 2)
e. (2x – 1) and (x – 3)

Answer:

Which is a factor is x + 1 -> x = -1

f(-1) = 0
2(-1)³ – 5(-1)³ – p(-1) + 3 = 0
-2 – 5 + p + 3 = 0
p = 4

Then, f (x) = 2x³ -5x³ – 4x =3

= (x + 1)(2×2 – 7x + 3)
= (x + 1)(2x – 1)(x – 3)

So, the other factors are (2x – 1) and also (x – 3).

Answer: E

Polynomial: Definition, Value, Terms, Distribution and Example Problems

Problem 5.

There are Two polynomials x³ -4x³ – 5x + m and x² -3x – 2 ÷ x + 1 will have the same remainder, so 2m + 5 = …

a. 17
b. 18
c. 24
d. 27
e. 30

Answer:

For example f(x) = x³ -4x² – 5x + m and x² -3x – 2

If ÷(x + 1 ) –> x = -1 will have the same remainder, then:
f(-1) = g(-1)
(-1)³ – 4(-1)² + 5(-1) + m = (-1)² + 3(-1) – 2
-1 -4 – 5 + m = 1 – 3 – 2
-10 + m = -4
m = -4 + 10
m = 6

So, the value of 2m + 5 = 2(6) + 5 = 17

Answer: A

Thus the review from About the knowledge.co.id about Polynomial , hopefully can add to your insight and knowledge. Thank you for visiting and don't forget to read other articles.

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