Remainder and factor theorem

Remainder theorem: The remainder theorem states that if a polynomial function $p(x)$, of degree $n\ge 1$ is divided by $(x-a)$, then the remainder obtained is equal to $f(a)$. In other words, if we divide $p(x)$ by $(x-a)$ and obtain a quotient $q(x)$ and a remainder $R$, then $p(x) = (x-a)q(x) + R$, and $p(a) = R$. Here the degree of $q(x)$ is less than or equal to the degree of $p(x)$. ”This theorem is proved in the derivation section.”

In other words, the remainder theorem can be used to find the remainder when a polynomial is divided by a linear factor. This can be helpful in simplifying complicated expressions, and in solving problems that involve polynomials. Let's have a look at a few examples.

Example: Find the remainder when $p(x) = 3x^3 - 4x^2 + 2x - 1$ is divided by $(x-2)$.

Solution: To find the remainder, we can use the remainder theorem, which states that the remainder is equal to $p(2)$.

So, we evaluate $p(2)$ as follows:

$p(2) = 3(2)^3 - 4(2)^2 + 2(2) - 1$

$\;\;\;\;\;\;\;= 24 - 16 + 4 - 1$

$\;\;\;\;\;\;\;= 11$

Therefore, the remainder when $p(x)$ is divided by $(x-2)$ is $11$.

Example: Find the remainder when $p(x) = 2x^4 - 5x^3 + 4x^2 - 3x + 1$ is divided by $(x+1)$.

Solution: To find the remainder when $p(x) = 2x^4 - 5x^3 + 4x^2 - 3x + 1$ is divided by $(x+1)$, we can use the remainder theorem, which states that the remainder is equal to $p(-1)$. So, we evaluate $p(-1)$ as follows:

$p(−1)=2(−1)^4−5(−1)^3+4(−1)^2−3(−1)+1$

$\;\;\;\;\;\;\;\;\;\;\;=2+5+4+3+1$

$\;\;\;\;\;\;\;\;\;\;\;= 15$

Therefore, the remainder when $p(x)$ is divided by $(x+1)$ is $15$.

Zero of a polynomial: A zero of a polynomial is a value of the variable (usually denoted by x) that makes the polynomial equal to zero. In other words, a zero of a polynomial $p(x)$ is a value of x for which $p(x) = 0$.

For example, the polynomial $p(x) = x^2 - 5x + 6$ has two zeros: $x = 2$ and $x = 3$, because:

$p(2) = 2^2 - 5(2) + 6 = 0$

$p(3) = 3^2 - 5(3) + 6 = 0$

So, $2$ and $3$ are zeros of $p(x)$.

Factor Theorem: The Factor Theorem is a theorem that helps us to determine whether a given polynomial has a certain factor or not. The theorem states that if a polynomial $p(x)$ is divided by $(x-a)$, then the remainder is zero if and only if $(x-a)$ is a factor of $p(x)$. In other words, if $p(a) = 0$, then $(x-a)$ is a factor of $p(x)$. “This Theorem is proved in the derivation section.”

Example: Let's consider the polynomial $p(x) = x^2 - 5x + 6$. We can check whether $(x-2)$ and $(x-3)$ are factors of $p(x)$ using the Factor Theorem: 1. $(x-2)$ is a factor of $p(x)$ if and only if $p(2) = 0$:

$p(2) = 2^2 - 5(2) + 6 = 0$

Since $p(2) = 0$, we can conclude that $(x-2)$ is a factor of $p(x)$.

$(x-3)$ is a factor of $p(x)$ if and only if $p(3) = 0$:

$p(3) = 3^2 - 5(3) + 6 = 0$

Since $p(3) = 0$, we can conclude that $(x-3)$ is a factor of $p(x)$.

Therefore, we can factor $p(x)$ as follows:

$p(x) = x^2 - 5x + 6 = (x-2)(x-3)$

Example: Consider the polynomial $q(x) = x^3 - 2x^2 - x + 2$.

We can check whether $(x-1)$ is a factor of $q(x)$ using the Factor Theorem:

$(x-1)$ is a factor of $q(x)$ if and only if $q(1) = 0$:

$q(1) = 1^3 - 2(1^2) - 1 + 2 = 0$

Since $q(1) = 0$,

we can conclude that $(x-1)$ is a factor of $q(x)$

Derivations:

Remainder Theorem: The remainder theorem states that if a polynomial function $p(x)$, of degree $n\ge 1$ is divided by $(x-a)$, then the remainder obtained is equal to $f(a)$. In other words, if we divide $p(x)$ by $(x-a)$ and obtain a quotient $q(x)$ and a remainder $R$, then $p(x) = (x-a)q(x) + R$, and $p(a) = R$. Here the degree of $q(x)$ is less than or equal to the degree of $p(x)$.

Proof:

Let $p(x)$ be a polynomial of degree $n$, and let $g(x)$ be the polynomial $x-a$. We want to find the remainder $R$ when $p(x)$ is divided by $g(x)$.

Using the polynomial division algorithm, we can write:

$p(x) = q(x) g(x) + R$ where $q(x)$ is the quotient and $R$ is the remainder.

Since $g(x)$ is of degree $1$, we can write:

$g(x) = x - a$

Using this, we can rewrite the above equation as:

$p(x) = q(x) (x-a) + R$

Now, if we substitute $x = a$ into this equation, we get:

$p(a) = q(a) (a-a) + R$

$p(a) = R$

Therefore, the remainder $R$ when $p(x)$ is divided by $x-a$ is equal to $p(a)$. This completes the proof of the Remainder Theorem.

Factor Theorem: The Factor Theorem is a theorem that helps us to determine whether a given polynomial has a certain factor or not. The theorem states that if a polynomial $p(x)$ is divided by $(x-a)$, then the remainder is zero if and only if $(x-a)$ is a factor of $p(x)$.

Proof:

Assume that $p(x)$ has a factor $(x-a)$. Then we can write:

$p(x) = (x-a)g(x)$

where $g(x)$ is a polynomial of degree $n-1$.

Now, if we substitute $x = a$ into this equation, we get:

$p(a) = (a-a)g(a) = 0$

Therefore, $p(a) = 0$, which confirms the "only if" part of the Factor Theorem.

Conversely, assume that $p(a) = 0$. We want to prove that $p(x)$ has a factor $(x-a)$. To do this, we will use the Remainder Theorem. If we divide $p(x)$ by $(x-a)$, we get:

$p(x) = (x-a)q(x) + R$ where $q(x)$ is the quotient and $R$ is the remainder. Now, if we substitute $x = a$ into this equation, we get:

$p(a) = (a-a)q(a) + R$

$p(a) = R$

Since we assumed that $p(a) = 0$, it follows that $R = 0$. Therefore, $p(x)$ is exactly divisible by $(x-a)$, which proves the "if" part of the Factor Theorem.

Therefore, we have proven that a polynomial $p(x)$ has a factor $(x-a)$ if and only if $p(a) = 0$.