Operations with Polynomials

Add the polynomials$3x^3+2x^2+6x+5$ and$x^4+x^2+3x+2$.

The resulting polynomial is:

$0x^4+3x^3+2x^2+6x+5$

$x^4+0x^3-x^2+3x+2$ +

$\mathbf{}{\mathit{x}}^{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{9}\mathit{x}\mathbf{}\mathbf{+}\mathbf{7}$

Subtract the polynomials$3x^3+2x^2+6x+5$ and $x^4+x^2+3x+2$.

Horizontal method

$(3x^3+2x^2+6x+5)-(x^4-x^2+3x+2)$

$3x^3+2x^2+6x+5-x^4+x^2-3x-2$ Change the signs of all the terms in the second polynomial.

$3x^3+\mathbf{2}{\mathit{x}}^{\mathbf{2}}+\mathbf{6}\mathit{x}+\mathbf{5}-x^4+{\mathit{x}}^{\mathbf{2}}-\mathbf{3}\mathit{x}-\mathbf{2}$ Combine like terms.

$\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}{\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathit{x}\mathbf{}\mathbf{+}\mathbf{3}$

Vertical method

$\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{3}$ Add like terms.

Multiply $2x^2-3x+5$ and $(x+2)$ using the horizontal and vertical methods.

Horizontal method

Multiply each term in the first polynomial by the second polynomial (distributive property)

Expand brackets and combine like terms

Vertical method

______________

$4x^2-6x+10$ Multiply $2x^2-3x+5$ by $2$

$2x^3-3x^2+5x$ Multiply $2x^2-3x+5$ by $x$

______________

$\mathbf{2}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{}\mathit{x}\mathbf{+}\mathbf{10}$ Add like terms

Divide$x^3+x^2-36$ by$\left(x-3\right)$ using long division.

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.

1. 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.
2. Multiply the first term of the quotient$x^2$ by both terms in the divisor, and place the results under the dividend lined up with their corresponding exponent.
3. Subtract like terms, being careful with the signs.
4. Bring down the next term in the polynomial (dividend).
5. Repeat steps 1 - 4 until the degree of the expression in the remainder is lower than the one of the divisor.

Complete missing terms with coefficient zero.

Subtract like terms.

Bring down 0x.

Subtract like terms.

Bring down –36.

Subtract like terms.

The remainder is zero.

The result can be expressed in the following way: $\frac{x^3+x^2-36}{x-3}=x^2+4x+12$.

Divide $x^3+x^2-36$ by $(x-3)$ using synthetic division.

• The divisor is $(x-3)$, therefore we evaluate the dividend when $x=3$.

• Bring down the leading coefficient below the horizontal line. Multiply the leading coefficient by the value of x. Write down the result of the multiplication just under the following coefficient. Then, add the values in the second column taking into account their signs.

• Multiply the result of the addition by the value of x, and put the result of the multiplication just under the following coefficient. Then, add the values taking into account their signs. Repeat this step for all the coefficients.

The final value below the horizontal line will be the value of $f\left(3\right)$.

$f\left(3\right)=0$, therefore the remainder of the division is zero (0).

The result is: $\frac{x^3+x^2-36}{x-3}=x^2+4x+12$.

Factor $x^2+5x+6=0$.

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.

$(x+2)(x+3)=0$

Now to find the solutions that make the equation above equal to zero, we need to consider the zero product property.

So, we need to make each factor in $(x+2)(x+3)=0$ equal to zero, and solve for x.

$$\begin{gathered} x+2=0x+3=0 \\ x+2\mathbf{-}\mathbf{2}=\mathbf{-}\mathbf{2}x+3\mathbf{-}\mathbf{3}=\mathbf{-}\mathbf{3} \\ \boxed{\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{2}}\boxed{\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}} \end{gathered}$$

Factor $2x^2+13x+15=0$.

1. Multiply the coefficient of $x^2$ and the constant.

$2\times 15=30$

2. Find the factors of 30.

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.

$2x^2+3x+10x+15=0$

$2x^2+3x+10x+15=0$ Pull out common factors from both groups.

Make each factor in $\left(2x+3\right)\left(x+5\right)=0$ equal to zero, and solve for x.

$$\begin{gathered} 2x+3=0x+5=0 \\ 2x+3\mathbf{-}\mathbf{3}=\mathbf{-}\mathbf{3}x+5\mathbf{-}\mathbf{5}=\mathbf{-}\mathbf{5} \\ 2x=-3\boxed{\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{5}} \\ \frac{\cancel{2}x}{\cancel{2}}=\frac{-3}{2} \\ \boxed{\mathbf{x}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}} \end{gathered}$$

Factor $6x^3+11x^2+4x=0$.

$x(6x^2+11x+4)=0$

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

$6\times 4=24$

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

$x(6x^2+3x+8x+4)=0$ Now factor by grouping the expression inside the parentheses.

$x\left(3x\right(2x+1)+4(2x+1\left)\right)=0$

$x\left(\right(2x+1\left)\right(3x+4\left)\right)=0$

Make each factor in $x\left(\right(2x+1\left)\right(3x+4\left)\right)=0$ equal to zero, and solve for x.

$$\begin{gathered} \boxed{\mathbf{x}\mathbf{=}\mathbf{0}}2x+1=03x+4=0 \\ 2x+1\mathbf{-}\mathbf{1}=\mathbf{-}\mathbf{1}3x+4\mathbf{-}\mathbf{4}=\mathbf{-}\mathbf{4} \\ 2x=-13x=-4 \\ \frac{\cancel{2}x}{\cancel{2}}=\frac{-1}{2}\frac{\cancel{3}x}{\cancel{3}}=\frac{-4}{3} \\ \boxed{\mathbf{x}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\boxed{\mathbf{x}\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}}} \end{gathered}$$

The solutions are$x=0,x=\frac{-1}{2}$, and$-\frac{4}{3}$.

Simplify the following fractions:

• $\frac{\left(x+4\right)\left(3x-1\right)}{\left(3x-1\right)}=\frac{\left(x+4\right)\cancel{\left(3x-1\right)}}{\cancel{\left(3x-1\right)}}=x+4$ Cancel the common factor $(3x-1)$.

• $\frac{x^2+x-12}{(x-3)}$ Factor the numerator.

$\frac{x^2+x-12}{(x-3)}=\frac{\cancel{\left(x-3\right)}\left(x+4\right)}{\cancel{\left(x-3\right)}}=x+4$ Cancel the common factor $(x-3)$.

• $\frac{x^2+3x+2}{x^2+5x+4}$ Factor the numerator and denominator.

$\frac{x^2+3x+2}{x^2+5x+4}=\frac{\cancel{\left(x+1\right)}\left(x+2\right)}{\cancel{\left(x+1\right)}(x+4)}=\frac{x+2}{x+4}$ Cancel the common factor $(x+1)$.

• $\frac{2x^2+7x+6}{(x-5)(x+2)}$ Factor the numerator.

$2\times 6=12$

The factors of 12 that give 7 when added together are 3 and 4.

$2x^2+3x+4x+6=0$ Now factor by grouping and pull out common factors.

$x(2x+3)+2(2x+3)=0$

$(2x+3)(x+2)=0$

Going back to the fraction:

$\frac{2x^2+7x+6}{(x-5)(x+2)}=\frac{\left(2x+3\right)\cancel{\left(x+2\right)}}{\left(x-5\right)\cancel{\left(x+2\right)}}=\frac{2x+3}{x-5}$ Canceling the common factor$(x+2)$.

• Show that$(x-1)$ is a factor of $4x^3-3x^2-1$.