Introduction to Polynomials

A polynomial is a function defined for all real x involving positive powers of x in the form: $P(x)={ p }_{ 0 }+{ p }_{ 1 }x+{ p }_{ 2 }{ x }^{ 2 }+{ p }_{ 3 }{ x }^{ 3 }+...+{ p }_{ n-1 }{ x }^{ n-1 }+{ p }_{ n }{ x }^{ n }$

...assuming that P(x) is a continuous and differentiable function.

The above polynomial expression has degree n where ${ p }_{ n }\neq 0$. While the p's are called coefficients. ${ p }_{ n }{ x }^{ n }$ is called the leading term and ${ p }_{ n }$ the leading coefficient, while ${ p }_{ 0 }$ is called the constant term.

If ${ p }_{ n }=1$P(x) is called a monic polynomial.

While if ${ p }_{ 0 }={ p }_{ 1 }={ p }_{ 2 }=...={ p }_{ n }=0$ then P(x) is the zero polynomial.

Examples

The real values of x that satisfy the equation are called the real roots of the equation or the real zeroes of the polynomial.

Examples

1. Show that the polynomial $p(x)={ x }^{ 2 }-x+4$ has no real zeroes. $p(x{ x }^{ 2 }-x+4=0$

Use discriminant: ${ b }^{ 2 }-4ac={ (-1) }^{ 2 }-4(1)(4)\\ \qquad =1-16\\ \qquad =-15\qquad <0$

So the polynomial has no real zeroes.

2. For the polynomial $P(x)=a{ x }^{ 5 }-3{ x }^{ 4 }+{ x }^{ 3 }-7x+1$

a) Evaluate a if the polynomial has real zeroes.

b) Find the degree of the derivative $P'(x)$

a) For monic polynomial, a = 1

b) $P'(x)=5{ x }^{ 4 }-12{ x }^{ 3 }+3{ x }^{ 2 }-7$ $P'(x)$ has degree 4 (highest power).

Division of Polynomials

A polynomial P(x) can be written as $P(x)=A(x)\cdot Q(x)+R(x)$ where P(x) is the dividend, A(x) is the divisor, Q(x) is the quotient, and R(x) is the remainder.

The degree of remainder R(x) is always less than the degree of the divisor A(x).

For example, divide $5{ x }^{ 3 }-{ x }^{ 2 }+6$ by $x-4$. This means that the $P(x)=5{ x }^{ 3 }-{ x }^{ 2 }+6,\quad A(x)=x-4,\quad Q(x)=5{ x }^{ 2 }+19x+76,\quad R(x)=310$.

Synthetic Division of Polynomials is a shortcut method for dividing a polynomial by a linear factor. It is a condensed form of the long division process whenever a polynomial is divided by a first degree polynomial x - a.

For example, let us divide $2{ x }^{ 4 }+{ x }^{ 2 }-5x+2$, the leading coefficient is 2. Since the term ${ x }^{ 3 }$ is missing, which means its coefficient is zero.

Applying synthetic division as below: Multiply 3 by the divisor 2 giving us 6. We add 0 and get 6. The result is shown below on the right. An continuing the process, we get: The coefficient of the last row is 2, 6, 19, 52 and a remainder of 158.

Therefore, the result will be: $2{ x }^{ 4 }+6{ x }^{ 3 }+19{ x }^{ 2 }+52x-+158$.

Another example: Remainder Theorem

If a polynomial P(x) is divided by x - a, then the remainder is P(a).

Examples

1. Find the remainder when $3{ x }^{ 4 }-2{ x }^{ 2 }+5x+1$ is divided by x - 2.

By remainder theorem, $P(2)=3{ (2) }^{ 4 }-2{ (2) }^{ 2 }+5(2)+1\\ \qquad =\quad 51$

So the remainder is 51.

2. Evaluate m if the remainder is 4 when dividing $2{ x }^{ 4 }+mx+5$ by x + 3.

The remainder when P(x) is divided by x + 3 is P(-3). So, $P(-3)=4\\ 2{ (-3) }^{ 4 }+m(-3)+5=4\\ 162-3m+5=4\\ 3m=163\\ \\ m=54\cfrac { 1 }{ 3 }$

From these two examples then we can say:

For polynomial P(x), if P(a) = 0 then x - a is a factor of the polynomial.

And polynomial P(x), if x - a i a factor of the polynomial, then P(a) = 0.

More Properties of Polynomial

If polynomial P(x) has degree n and n distinct zeroes ${ a }_{ 1 },\quad { a }_{ 2 },\quad { a }_{ 3 },\quad ...,\quad { a }_{ n }$, then $P(x)={ p }_{ n }(x-{ a }_{ 1 })(x-{ a }_{ 2 })(x-{ a }_{ 3 })...(x-{ a }_{ n })$.

A polynomial of degree n cannot have more than n distinct real zeroes.

If two polynomials of degree n are equal for more than n distinct values of x, then the coefficients of like powers of x are equal. That is: ${ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+...+{ a }_{ n }{ x }^{ 3 }\equiv { b }_{ 0 }+{ b }_{ 1 }x+{ b }_{ 2 }{ x }^{ 2 }+...+{ b }_{ n }{ x }^{ 3 }$

If x - a is a factor of polynomial P(x), then a is a factor of the constant term of the polynomial.

Examples

1. Find all factors of $f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }-4x-12$.

Try factors of -12 (i.e. $\pm 1,\quad \pm 2,\quad \pm 3,\quad \pm 4,\quad \pm 6,\quad \pm 12p/latex])

f\left( 2 \right) ={ (2) }^{ 3 }+3{ (2) }^{ 2 }-4(2)-12\\ \qquad =\quad 0$

Therefore x - 2 is a factor of f(x). We can use this to find other factors: Therefore, $f\left( x \right) =(x-2)({ x }^{ 2 }+5x+6)\\ \qquad =(x-2)(x+2)(x+3)$

2. Write ${ x }^{ 3 }-2{ x }^{ 2 }+5$ in the form $a{ x }^{ 3 }+b{ \left( x+3 \right) }^{ 2 }+c(x+3)+d$.