# Polynomials & Cubic Equations

•  Introduction to Polynomials.
•  Equality of Polynomials.

## Introduction to Polynomials

A polynomial in $x$ is an algebraic expression consisting of terms with non-negative integer powers of $x$ only.

An equation will not be considered a polynomial if the power is negative or fraction. If the variable $x$ is in the denominator or in the root, it is not considered a polynomial.

Some of the examples of polynomials and non-polynomials are listed below.

 Polynomials Non-Polynomials \begin{align*} 2x^3-x+3 \end{align*} \begin{align*} 5x^2-3x^{-1}+1 \end{align*} \begin{align*} x^7-x^3-5 \end{align*} \begin{align*} x^2-x^{^\tfrac {1}{2}}+3 \end{align*} \begin{align*} 2+\frac{2}{5}x^3 \end{align*} \begin{align*} \frac{2}{x}+5 \end{align*} $7$ \begin{align*} 2\sqrt{x}+x-7 \end{align*}

## Degree of Polynomial

The degree of a polynomial is the highest power of its individual terms with non-zero coefficients.

 Polynomial Degree \begin{align*} 2x^3-x+3 \end{align*} $2$ \begin{align*} x^7-x^3-5 \end{align*} $7$ \begin{align*} 2+\frac{2}{5}x^3 \end{align*} $3$

In the above table, you can see that the degree of any polynomial would be the highest power in the equation.

If the equation looks as follows, where there is no variable, then we need to consider it as $x^0$. Because the value of  $x^0$ is $1$. Therefore the degree would be $0$.

 Polynomial Degree $7x^0$ $0$

## Evaluating Polynomials

To denote polynomials, we can use $P$ and $Q$ instead of $f$ and $g$. Polynomials should be denoted only using Capital letters.

The symbol to denote polynomial is $P(x)$ if the equation has $x$ as variable.

Example:

\begin{align*} P(x) = 2x^2 -x+3 \end{align*}

The $x$ in $P(x)$ represents the variable $x$.

If the variable of the equation is $y$, the symbol to denote the polynomial is $P(y)$

For example, \begin{align*} P(y) = 2y^2 -y+3 \end{align*}

The $y$ in $P(y)$ represents the variable $y$.

Example 1:

It is given that $P(x) = 2x^2 - x + 3$ and \begin{align*} Q(x) = x^3 + 2x -1 \end{align*}

Evaluate $P(2) +Q(-1)$.

Solution:

To find $P(2)$, we need to replace the $x$ with $2$.

\begin{align*} P(2) = 2(2)^2 - 2 +3 \end{align*}

To find $Q(-1)$, we need to replace the $x$ with $-1$.

\begin{align*} Q(-1) = (-1)^3 + 2(-1) -1 \end{align*}

Adding both of the above equations.

\begin{align*} P(2) +Q(-1) = [\;2(2)^2 - 2 +3\;] \;\;+\;\;[\;(-1)^3 + 2(-1) -1\;] \end{align*}

Solving the above will give the answer.

$=5$

Therefore, the answer is $5$.

Example 2:

Given that \begin{align*} x^3 - 5x+3 = (Ax+3)(x-1)(x-2) \;+\; B(x-1) \;+\; C \end{align*} for all real values of $x$, find the values of $A,B$, and $C$.

Solution:

In the question, it is given as “for all real values of $x$” which means we can substitute any value for $x$ and it will be correct.

To find C:

One needs to carefully think and choose an $x$ value so that the other terms like $A$ and $B$ will become Zero.

When we substitute $x=1$ except for $C$, all other terms on the R.H.S. will become Zero so that the value of $C$ can be found.

\begin{align*} \big(1\big)^3 - 5\big(1\big)+3 = \big(\;A(1)+3\;\big)\big(\;1-1\;\big)\big(\;1-2\;\big)+B\big(\;1-1\;\big) + C \end{align*}

Solving the above equation will give,

\begin{align*} & -1 = 0+0 + C \\ & C = −1 \end{align*}

Therefore,    $C= −1.$

To find B:

Substitute $x=2$, and the value of $C$ to get the value of $B$. Thus the term $A$ becomes Zero. So that you can find $B$ easily.

\begin{align*} \big(2\big)^3 - 5\big(2\big)+3 = \big(A(2)+3\big)\big(2-1\big)\big(2-2\big)+B\big(2-1\big) + C \end{align*}

Solving the above equation will give,

\begin{align*} & 1 = 0+B(1) -1 \\ & 1 + 1 = B \\ & B = 2 \end{align*}

Therefore, $B=2.$

To find A:

Substitute $x=3$, and the value of $C, B$ to get the value of $A.$

\begin{align*} & (3)^3 - 5(3)+3 = [\;A(3)+3\;] (3-1) (3-2) +2(3-1) -1 \\ & 15 = (3A+3)(2)(1)+4 -1 \\ &15 = (3A+3)(2)+3 \\ &15 -3= (3A+3)(2) \\ & 12 = (3A+3)(2) \\ & 6 = (3A+3) \\ & 3 = 3A \\ &A =1 \end{align*}

Therefore, $A=1.$

The answer is \begin{align*} A=1, \;B=2, \;C=−1 \end{align*}.

Question 1:

If \begin{align*} 4x^3 + x^2 -26x+12=(Ax-3)(x-2)(x+3) + B(x-2)+C \end{align*} for all real values of $x$, find the values of $A,B$, and $C$.

Solution:

In the question, it is given as “for all real values of $x$” which means we can substitute any value for $x$ and it will be correct.

Substituting $x=2$, to find $C$, the term $A$ and $B$ becomes $0$.

\begin{align*} & 4(2)^3 +(2)^2 -26(2)+12 =[\;A(2)-3\;] [\;(2)-2\;][\;(2)+3\;] + B[\;(2)-2\;]+C \\ & 32+4 -52+12 =0+0+C \\ & -4 =0+0+C \\ & C =-4 \end{align*}

Substituting $x=-3$ and $C$, to find $B$. The term $A$ becomes $0$.

\begin{align*} & 4(-3)^3+ (-3)^2-26(-3)+12 &=[\;A(-3)-3\;][\;(-3)-2\;)[\;(-3)+3\;] + B[\;(-3)-2\;]-4 \\ & -9 =-5B-4 \\ & -5 =-5B \\ & B =1 \end{align*}

Therefore, $B=1$.

Substitute $x=5$ and $C$ and $B$ to find $A$.

\begin{align*} & 4(5)^3 + (5)^2 -26(5)+12 = [\;A(5)-3\;][\;(5)-2\;][\;(5)+3\;] + (1)[\;(5)-2\;]+(-4) \\ & 407 = (5A - 3) (3) (8) + 3-4 \\ & 407 = (5A - 3) 24 -1\\ & 408 = (5A - 3) 24\\ &17 = (5A - 3)\\ & 20 = 5A \\ & A =4 \end{align*}

Therefore, $A=4$.

The correct answer is $A=4,\;B=1,\;C=-4$.

Question 2:

Given that \begin{align*} 2x^2 + x + C = A(x+1)^2 +B(x+1)+4 \end{align*} for all real values of $x$, find the values of $A,B$, and $C$.

Solution:

In the question, it is given as “for all real values of $x$” which means we can substitute any value for $x$ and it will be correct.

Substituting $x= -1$, the terms $A$ and $B$ become $0$.

\begin{align} & 2(-1)^2 -1+C = 4 \\ & C = 3\\ \end{align}

Therefore, $C =3$.

But to find $A$ and $B$, we cannot substitute number $​​-1$, because if we do this, then both the $A$ and $B$ terms will be $0$. Hence, we need to follow a method called comparing coefficients. The method of comparing coefficients is applicable only if the equation is Polynomial.

\begin{align} 2x^2 + x + C = A(x+1)^2 +B(x+1)+4 \end{align}

The term $(x+1)^2$ can be expanded according to the formula \begin{align} (a+b)^2 = a^2 +2ab+ b^2 \end{align}

Therefore,

\begin{align} 2x^2 + x + C &= A(x^2 +2x+1)+Bx+B+4\\ 2x^2 + x + C &= Ax^2 +2Ax+A+Bx+B+4\\ 2x^2 + x + C &= Ax^2 +2Ax+Bx+A+B+4\\ 2x^2 + x + C &= Ax^2 +(2A+B)x+A+B+4\\ \end{align}

Compare the coefficients of $x^2$ to get the value of $A$.

Therefore $A=2$.

Compare the coefficients of $x$ to get the value of $B$.

$2A + B = 1$

Substituting the value of $A$.

\begin{align} & 2(2) + B = 1\\ & 4 + B = 1\\ & B = -3\\ \end{align}

Therefore, $B= -3$.

The other method to find $B$ is by comparing the last terms of the $L.H.S.$ and $R.H.S.$

\begin{align} & A+B+4 = C \\ & 2−3+4 = C \\ & C= 3\\ \end{align}

The answers are $A=2, \;B = -3, \;C =3$.

## Conclusion

In this article, we learnt about Polynomials and Cubic equations as per the Secondary 3 Additional Mathematics level in Singapore. We learnt about the degree of the polynomial and how to solve them by substituting values of $x$ and by comparing coefficients. Practise these questions regularly to grab the topic efficiently.

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