\( f(x) = 3x^2 + 2x + 5\) this is the same as \( f(x) = 3x^2 + 2x^1 + 5x^0\)

From the **Exponential Rules** , remember that \(x^1 = x\) and \(x^0 = 1\).

**Polynomial Function Form**.

## Polynomial Function

The Polynomial Function follows the standard form:

\[ f(x) = a_n x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \dots + a_1 x + a_0 .\]

Notice that we write the terms in a polynomial in **decreasing order**, from the biggest exponent to the smallest.

The highest power or exponent present in a polynomial is called the **degree **of the polynomial.

Not all terms must be present in a polynomial. If any term is missing, you can assume that it has a coefficient of zero (0).

For example, if you have \(3x^2 + 8 = 0\), it is the same as \(3x^2 + 0x + 8 = 0\)

## Evaluating polynomials

To evaluate a polynomial, all you need to do is substitute \(x\) with a number to find its solution.

Evaluate \( f(x) = 3x^2 + 2x + 5\) when \(x=4\).

\[ \begin{align} f(4) &= 3\cdot 4^2 + 2\cdot 4 + 5 \\ &= 3\cdot 16 + 8 + 5 \\ &= 48 + 13 \\ &= 61 \end{align} \]

## Adding or subtracting polynomials

To **add polynomials,** you can use the distributive property to combine like terms so that you end up with only one term for each exponent, like this:

Add \( 3x^3 + 2x^2 + 6x + 5\) and \(x^4 + x^2 + 3x + 2\)

Combining like terms,

\[ \begin{align} (3x^3 + 2x^2 + 6x + 5 )+ (x^4 + x^2 + 3x + 2 ) &= x^4 + 3x^3 + (2x^2 + x^2) + (6x + 3x) + (5+2) \\ &= x^4 + 3x^3 + 3x^2 + 9x + 7.\end{align} \]

Another method is to stack the polynomials one on top of the other so that you can see the like terms more easily. In this case, you need to complete the missing terms assuming that they have a coefficient of zero (0):

\[ \begin{array}{ll} &0x^4 + 3x^3 + 2x^2 + 6x + 5 \\ + & \phantom{0}x^4 + 0x^3 + \phantom{0} x^2 + 3x + 2 \\ \hline & \phantom{0} x^4 + 3x^3 + 3x^2 + 9x + 7 \end{array} \]

You can see that the result is the same as before, but this method gives you a more organized way to identify like terms and avoids confusion.

To **subtract polynomials,** you can follow the same method, but remember to be careful with the signs. Now we will subtract the same polynomials as before:

\[ \begin{array}{ll} &0x^4 + 3x^3 + 2x^2 + 6x + 5 \\ - & (\phantom{0}x^4 + 0x^3 + \phantom{0} x^2 + 3x + 2) \end{array} \]

is the same as

\[ \begin{array}{ll} &0x^4 + 3x^3 + 2x^2 + 6x + 5 \\ + & -x^4 - 0x^3 - \phantom{0} x^2 - 3x - 2 \end{array} \]

so

\[ \begin{array}{ll} &0x^4 + 3x^3 + 2x^2 + 6x + 5 \\+& -x^4 + 0x^3 - \phantom{0} x^2 - 3x - 2 \\ \hline & -x^4 + x^3 + 3x^2 + 9x + 7 \end{array} \]

## How to Factorize Polynomials?

**Factoring polynomials **involves rewriting polynomials as the product of two or more simpler terms. You need to approach these problems differently, depending on the degree of the polynomial and the coefficient of the term with the highest power:

**If the degree is 2 and the coefficient of \(x^2\) is 1**, you simply find the factors that make the polynomial equal to zero.

\(x^2+5x+6=0\)

Find the factors of 6 that when multiplied, equal +6 and when added equal +5. In this case, the factors that match the requirements are 2 and 3.

\[ (x+2)(x+3)=0\]

**If the degree is 2 and the coefficient of \(x^2\)****is not 1**: Here, you have a few more steps to follow:

\(2x^2+13x+15 = 0\)

1. Multiply the coefficient of \(x^2\) and the constant.

\(2 \cdot 15 = 30\)

2. Find the factors of 30.

If the sign of the constant is positive, you need to add the factors of 30 that give the coefficient of x when added together. If the sign of the constant is negative, you need to include the factors of 30 that give the coefficient of x when subtracted.

1 | |

2 | |

3 | |

5 |

The factors of 30 that give 13 when added together are 3 and 10.

3. Copy the \(2x^2\) term and the constant as in the original polynomial, and in between these terms, add the factors found in the previous step.

\[2x^2 + 3x+10+15=0\]

4.**Factor by grouping**the first two terms \(2x^2+3x\) and the last two terms \(10x+15\)

Pull out the common factor from both groups:

\[ 2x^2 + 3x + 10x + 15 = 0 \]

Now that the terms in the parentheses match, take out the common factor:

\[\begin{align} x(2x+3) + 5(2x+3) &= 0 \\ (2x+3)(x+5)&=0 \end{align} \]

The two solutions are \( x = -\frac{3}{2}\) and \(x=-5\).

**If the degree is greater than 2:**In this case, you might need to pull out common factors, if possible, and also use factor by grouping.

In the equation \(6x^3 + 11x^2 + 4x=0\), \(x\) is a common factor so pull it out first to get

\[ x(6x^2 + 11x + 4) = 0\]

then follow the steps in the previous example for \(6x^2 + 11x + 4\).

Now follow the steps from the previous case when the degree is 2, and the coefficient of \(x^2\) is not 1.

\(6 \cdot 4 = 24\)

24 | |

The factors of 24 that give 11 when added together are 3 and 8. So

\[ \begin{align} 6x^2 + 11x + 4 &= 3x(2x+1) + 4(2x+1) \\ &= (2x+1)(3x+4). \end{align}\]

That means

\[ \begin{align} 6x^3 + 11x^2 + 4x &= x(6x^2 + 11x + 4) \\ &= x(2x+1)(3x+4) \end{align} .\]

Looking at

\[x(2x+1)(3x+4) = 0 \]

the solutions are \(x=0\), \(x = -\frac{1}{2}\), and \(x = -\frac{4}{3}\).

## How can you simplify polynomials?

To simplify fractional algebraic expressions containing polynomials, you need to **factorise** the numerator and denominator, then **cancel common factors**.

Simplify the following fractions:

- Canceling the common factor \(3x-1\), \[ \frac{(x+4)(3x-1)}{3x-1} = \frac{(x+4) \cancel{(3x-1)}}{\cancel{3x-1}} = x+4 \]

Factoring the numerator and then canceling the common factor \(x-3\) \[ \frac{x^2 + x - 12}{x-3} = \frac{(x+4)(3x-1)}{3x-1} = \frac{(x+4) \cancel{(3x-1)}}{\cancel{3x-1}} = x+4 \]

Factoring the numerator and denominator and then canceling the common factor \(x+1\) \[ \begin{align} \frac{x^2 + 3x + 2}{x^2 + 5x + 4} &= \frac{ (x+1)(x+2)}{(x+1)(x+4)} \\ &= \frac{ \cancel{(x+1)}(x+2)}{\cancel{(x+1)}(x+4)} \\ &= \frac{x+2}{x+4}. \end{align}\]

Factoring the numerator and denominator and then cancelling the common factor \(x+5\)\[ \begin{align} \frac{2x^2+13x+15}{x^2 + 4x - 5} &= \frac{(2x+3)(x+5)}{(x-1)(x+5)} \\ &= \frac{(2x+3)\cancel{(x+5)}}{(x-1)\cancel{(x+5)}} \\ &= \frac{2x+3}{x-1}. \end{align}\]

## Dividing polynomials

To divide polynomials, you use the long division method. Here's an example of the steps to follow:

Divide \(x^3 + x^2 -36\) by \(x-3\)

Before we start, we need to identify each part of the division. \(x^3 + x^2 -36\) is the **dividend**, \(x-3\) is the **divisor**, and the result is called the **quotient**. Whatever is left at the end is the **remainder**.

Remember to complete any missing terms in the dividend with coefficient zero so that the polynomial is in decreasing order of exponents.

First of all, divide the first term of the dividend \(x^3\) by the first term of the divisor \(x\), then put the result \(x^2\) as the first term of the quotient.

Multiply the first term of the quotient \(x^2\) by each term in the divisor, and place the results under the dividend lined up with their corresponding exponent.

Subtract like terms, being careful with the signs.

Bring down the next term in the polynomial (dividend).

Repeat steps 1 - 4 until the degree of the expression in the remainder is lower than the one of the divisor.

\[ \begin{array}{rll} x^2 + 4x + 12 && \\ x-3 \enclose{longdiv}{\; x^3 + x^2 + 0x - 36}\kern-.2ex \\ \underline{-\; (x^3 -3x^2)\phantom{+ 0x - 36 }} && \hbox{Subtract like terms} \\ 4x^2 + 0x \phantom{-360}&& \hbox{Bring down $0x$} \\ \underline{\phantom{ x^3 }-(4x^2-12x)\phantom{+0}} && \hbox{Subtract like terms} \\ \phantom{x^3 + x^2}12x-36 \phantom{0}&& \hbox{Bring down $-36$} \\ \underline{\phantom{ x^3 + x^2}-(12x-36 )} && \hbox{Subtract like terms} \\ \phantom{00}0 && \hbox{The remainder is zero.} \end{array}\]

## How to use the factor theorem with polynomials?

The factor theorem can be used to speed up the factoring process. It states that if you substitute a value \(p\) in a polynomial function \(f(x\) and it gives zero as a result \(f(p)=0\), then you can say that \(x - p\) is a factor of \(f(x)\)

It is especially useful in the case of cubic polynomials, and you can proceed as follows:

You can substitute values into \(f(x)\) until you find a value that makes\(f(p)=0\) .

Divide the\(f(x)\) by \(x - p\) as the remainder will be zero.

Write \(f(x)\) as \((x - p)g(x)\) where \(g(x)\) is a polynomial of smaller degree

Factor the remaining quadratic factor to write \(f(x)\) as the product of 3 linear factors.

- Show that \(x-1\) is a factor of \(4x^3 - 3x^2 - 1\)\[ \begin{align} f(1) &= 4(1)^3 - 3(1)^2 - 1 \\ &= 4-3-1 \\ &= 0 ,\end{align}\]

so \(x-1\) is indeed a factor of \(4x^3 - 3x^2 - 1\)

- Show that \(x-1\) is a factor of \(x^3 + 6x^2 + 5x - 12\) \[ \begin{align} f(1) &= (1)^3 +6(1)^2 +5(1) - 12 \\ &= 1+6+5-12 \\ &= 0 ,\end{align}\] so \(x-1\) is indeed a factor of \(x^3 + 6x^2 + 5x - 12\)

Divide the \(f (x)\) by \(x-p\)

\[ \begin{array}{rll} x^2 + 7x + 12 && \\ x-1 \enclose{longdiv}{\; x^3 + 6x^2 + 5x - 12}\kern-.2ex \\ \underline{-\; (x^3 -x^2)\phantom{+ 0x - 36 }} && \hbox{Subtract like terms} \\ 7x^2 + 5x \phantom{-360}&& \hbox{Bring down $5x$} \\ \underline{\phantom{ x^3 }-(7x^2-7x)\phantom{+0}} && \hbox{Subtract like terms} \\ \phantom{x^3 + x^2}12x-12 \phantom{0}&& \hbox{Bring down $-12$} \\ \underline{\phantom{ x^3 + x^2}-(12x-12 )} && \hbox{Subtract like terms} \\ \phantom{00}0 && \hbox{The remainder is zero.} \end{array}\]

Write \(f (x)\) as \((x-p)g(x)\) then factor \(g(x)\)\[ \begin{align} x^3 + 6x^2 + 5x - 12 &= (x-1)(x^2 + 7x + 12) \\ &= (x-1)(x+3)(x+4). \end{align} \]

## Polynomials - Key takeaways

Polynomials are expressions with multiple terms that contain a variable raised to a series of positive whole-number exponents, and each term may be multiplied by coefficients.

You write the terms in a polynomial in decreasing order of exponent.

To evaluate a polynomial, substitute x with a number to find its solution.

To add, subtract and divide polynomial expressions, complete any missing terms assuming that they have coefficient zero.

Factoring polynomials involves rewriting them as the product of two or more simpler terms.

To simplify fractional algebraic expressions containing polynomials, factorise the numerator and denominator, then cancel common factors.

To divide polynomials, use the long division method.

The factor theorem can be used to speed up the factoring process, especially in the case of cubic polynomials.

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##### Frequently Asked Questions about Polynomials

What is a polynomial?

Polynomials are expressions involving multiple terms that contain a variable raised to a series of positive whole-number exponents, each term may also be multiplied by coefficients.

How do you divide polynomials?

To divide polynomials follow these steps:

- First of all divide the first term of the dividend by the first term of the divisor, then put the result as the first term of the quotient.
- Multiply the first term of the quotient by each term in the divisor, and place the results under the dividend lined up with their corresponding exponent.
- Subtract like terms, being careful with the signs.
- Bring down the next term in the polynomial (dividend).
- Repeat steps 1 - 4 until the degree of the expression in the remainder is lower than the one of the divisor.

How do you factor polynomials?

Factoring polynomials involves rewriting polynomials as the product of two or more simpler terms. The factoring process depends on the degree of the polynomial and the coefficient of the term with the highest power.

How do you solve polynomials?

This depends on the degree of the highest exponent in the polynomial and what you are being asked to do. Normally, solving polynomials involves simplifying and factoring the polynomial to find its solution.

How do you multiply polynomials?

To multiply polynomials you need to multiply each term in the first polynomial by each of the terms in the second polynomial to remove the brackets, then combine like terms.

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