Poly-nomial means 'many names'.

A polynomial with one term (e.g. $y = x$) is a 'monomial'. Two terms (e.g. $y= x^2 + x$) is a binomial.

Three terms (e.g. $y= x^3 + x^2 + x$) is a trinomial, four terms 'quartic', five terms 'quintic'.

There is also a 'zero polynomial': $θ(x)=0$, where $f(x) + θ(x) = θ(x) + f(x) = f(x)$.

A polynomial equation is one of the form:

$a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} ..... a_{1}$

where $a_n$ .... $a_1$ and n can be any real or irrational number.

A polynomial is a function with x to two or more difference powers: e.g. $ax^3 + bx^2 + cx + d = 0$

Inequalities are functions that contain an algebraic expression that does not equal, but is expressed as being more or less (and possibly also equal to) than, zero or a constant: e.g. $ax^3 + bx^2 + cx + d < 0$

Polynomial inequalities may be solved graphically (also using a GDC) or by means of an algebraic method.

for any two polynomials $f$ and $g$ there are polynomials $m$ and $n$ which satisfy: $f(x)=g(x)m(x) + n(x)$, for all real values of $x$.

Use long division to divide $4x^4 + 2x^3 + 3x^2 + 2x - 5$ by:

$x^2+x+2$

Solution:

1st division ⇒ divisor $4x^2$ : subtract $4x^2(x^2+x+2) = 4x^4+4x^3+8x^2$

Remainder: $-2x^3 - 5x^2 + 2x$

2nd division ⇒ divisor $-2x$ : subtract $-2x(x^2+x+2)$ from $-2x^3 - 5x^2 + 2x$

Remainder: $-3x^2 + 6x - 5$

3rd division ⇒ divisor $-3$ : subtract $-3(x^2+x+2)$ from $-3x^2 + 6x - 5$

Remainder: $9x + 1$

Solution: $(x^2+x+2)(4x^2-2x-3) + (9x+1)$

$f(x) = ax^3+bx^2+cx+d$, where $a≠0$, is a polynomial of the third degree.

If $d=0$, the graph passes through the origin: when $x=0$, $y=0$. The value of $d$ is the vertical translation of the graph.

$c$ causes the graph to stretch in the y direction.

If $b$ is zero, the graph has no maxima or minima, only an inflexion point on the y-axis, at the value of $y=d$. If $b≠0$, the curve may have a maximum and a minimum, or it may have a flex (inflexion point only). The shape of the curve depends on the sign of the leading coefficients (a, b, c).

$ \{ \table ax+by=e; cx+dy=f$

$(x,y)= \({ed-fb}/{ad-bc} , {ab-ec}/{ad-bc} \)$ , $ad-bc≠0$

A polynomial

$$f(x) = a_nx^n+a_{n-1}x^{n-1}+ ... + a_2x^2+a_1x+a_0$$$a_k ∈ ℝ$, $k=0,1,2,3,..., n$, $a_n ≠ 0$,

has a factor $(x-p)$, $p ∈ ℝ$ if and only if $f(p) = 0$.

The number of positive real roots of a polynomial is equal to the number of sign changes of its coefficients, or an even number less. e.g. $f(x)=x^4+3x^3-3x^2+5x-4$ has 3 sign changes (+ to - to + to -), so the function has 3 roots or two roots (even number less).

Given a polynomial:

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+ ... + a_2x^2+a_1x+a_0$$$a_k ∈ ℤ$, $a_n ≠ 0$, and an integer $p$ such that $f(p) = 0$, then $p$ is a factor of $a_0$.

In addition, $p-q$ is also a factor of $f(p)$.

Find all the possible integer zeros of the polynomial: $f(x) = x^3 - 2x^2 - 8x + 12$.

Solution: find all possible factors of 12.

$p ∈ ${$±1, ±2, ±3, ±4, ±6, ±12$}

$f(1) = 1^3 - 2(1)^2 - 8(1) + 12 = 3$

$p-1$ is a factor of 3, so $p ∈ ${$±1, $±3$}

∴ $p ∈ ${$-2, 0, 2, 4$}

$p$ is a member of the set of intersection of the two sets: $p ∈ ${$-2, 2$}

Horner's algorithm utilises a system of synthetic division to find the remainder and coefficients of the quotient polynomial.

Divide $f(x) = 6x^3 + 12x^2 - 9x + 8$ by $g(x) = x+4$.

$x + 4$ ⇒ $r=f(-4)$

$f(x) = ((6x + 12)x - 9)x + 8$

$f(-4) = ((6.-4 + 12).-4 - 9).-4 + 8$

$= (-24 + 12).-4 - 9).-4 + 8$

$= (-12.-4 - 9).-4 + 8$

$= (48 - 9).-4 + 8$

$= (39).-4 + 8$

$= -156 + 8$

$= -148$

6 | 12 | -9 | 8 | |

-4 | -24 | 48 | -156 | |

6 | -12 | 39 | -148 |

Newton's Polynomial Method for solving a complex polynomial employs a system of iterations (divided differences) to determine ever more accurately the intersection of a function with the x-axis. This is a demonstration of the irrepressible power of limits.

The curve $y = f(x)$ has a tangent at value $(x_0 , f(x_0))$ which intersects the x-axis at $x = x_1$. Similarly, the tangent at $(x_1 , f(x_1))$ intersects the x-axis at $x = x_2$. The series $x_0, x_1, x_2, ...$ can continue towards a limit of $f(x) = 0$ at point α.

The function must have a non-zero derivative for the interval a-b. If the tangent is ever parallel (f'(x) = 0) to the axis, no solution will be found.

The tangent to the function f(x) at $(x_0 , f(x_0))$ is:

$$y - f(x_0) = f'(x_0)⋅(x - x_0)$$At $(x_1 , 0)$:

$$0 - f(x_0) = f'(x_0)⋅(x - x_0)$$Solving for $x_1$, we find:

$$-{{f(x_0)}/{f'(x_0)}} = x_1 - x_0$$ $$x_1 = x_0 -{{f(x_0)}/{f'(x_0)}}$$And all of the intersections can be expressed similarly:

$$x_2 = x_1 -{{f(x_1)}/{f'(x_1)}}$$In general, Newton's Series is:

$$x_{n+1} = x_n -{{f(x_n)}/{f'(x_n)}} , (n ∈ ℕ) $$This method for was published by the French mathematician Joseph-Louis Lagrange, 1736 - 1813. However, his work was based on pioneering work by a number of mathematicians, such as Edward Waring and Leonhard Euler. Newton had already invented an alternative method, which which does not involve a laborious recalculation of the entire interpolant, if a minor change is made to any of the points.

Interpolation is a process of creating a polynomial from a limited series of known points. It allows other points to be predicted with accuracy. The Lagrange Polynomial is the the polynomial with the least degree (power of x) which accurately determines the corresponding y-value for any x-value.

n + 1 given points:

$$P_0(x_0;y_0), P_1(x_1;y_1), ..., P_n(x_n;y_n)$$where $x_0 < x_1 < ... < x_n$

The polynomial of Lagrange for 1 = 0, 1, ..., n:

$$L_i(x) = {(x-x_0)(x-x_1)...(x-x_{i-1})(x-x_{i+1})...(x-x_n)}/{(x_i-x_0)(x_i-x_1)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_n)}$$This may be written in short form:

$$L_i(x) = Π↙{j=0}↙{j≠i}↖n {(x-x_j)}/{(x_i-x_j)}$$where Π is the product total of all the arguments.

$$L(x) = y_0L_0(x) + y_1L_1(x) + ... + y_nL_n(x) = Σy_iL_i(x)$$Content © Renewable-Media.com. All rights reserved. Created : September 14, 2014

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1916 - 2004

Maurice Wilkins, 1916 - 2004, molecular biologist, was 'the third man of the double helix', as his biography title declares. Born in New Zealand, but did most of his professional work in England, Wilkins shared the Nobel Prize with Crick and Watson.

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