An expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
The degree of a polynomial is the highest power of the variable in the expression.
Multiplying a Polynomial by a Constant
To multiply a polynomial by a constant, multiply the constant by each of the coefficients of the polynomial. This is sometimes called distributing the constant through the polynomial.
To multiply $3x^2 - 2x + 5$ by $4$, multiply each of the coefficients by $4$.
\$\$4 \cdot (3x^2 - 2x + 5) = 12x^2 - 8x + 20\$\$
Adding Polynomials
To add two polynomials, combine the like terms, which have the same variable raised to the same exponent.
To add the two polynomials $(x^2 - 5x + 6) + (2x^2 + 3x - 2)$, first combine the two $x^2$-terms, $x^2 + 2x^2 = 3x^2$. Then combine the two $x$-terms, $-5x + 3x = -2x$. Then combine the two constant terms $+6 - 2 = +4$. The sum is $3x^2 - 2x + 4$.
Subtracting Polynomials
Subtracting polynomials is more complicated than adding polynomials.
\$\$ (3x^2 + 5x - 1) - (x^2 - 2x + 3) \$\$
Since the $-$ sign can be thought of as a negative $1$ $(-1)$ and the coefficient of the $x^2$ in the second polynomial is really a $1$, this can be rewritten as:
\$\$ (3x^2 + 5x - 1) - 1 (1x^2 - 2x + 3) \$\$
Distribute the $-1$ through the second polynomial. The parentheses are no longer needed.
\$\$ 3x^2 + 5x - 1 - 1x^2 + 2x - 3 \$\$
Combine like terms.
\$\$ 2x^2 + 7x - 4 \$\$
Multiplying Polynomials
Multiplying two polynomials requires multiplying each combination of one term from the first polynomial with one term from the second polynomial and then combining all the products. The most common type of polynomial multiplication is when each of the polynomials has just two terms (called binomials). The four combinations can then be remembered with the word FOIL.
First terms in each expression: $5x \cdot 2x = +10x^2$ Outer terms in each expression: $5x \cdot - 4 = -20x$ Inner terms in each expression: $+3 \cdot 2x = +6x$ Last terms in each expression: $+3 \cdot -4 = -12$
\$\$ 10x^2 - 20x + 6x - 12 = 10x^2 - 14x - 12 \$\$
If one or both of the polynomials has more than two terms, then the FOIL method does not apply. Instead, get all the combinations by multiplying the first term in the polynomial on the left by all the terms in the polynomial on the right. Then multiply the second term in the polynomial on the left by all the terms in the polynomial on the right. Continue until the last term in the polynomial on the left has been multiplied by all the terms in the polynomial on the right.
Multiply $(x + 3)(2x^2 - 4x + 5)$
Solution: First multiply the $x$ in the binomial by each of the terms of $2x^2 - 4x + 5$. Then multiply the $+3$ in the binomial by each of the terms of $2x^2 - 4x + 5$. Combine the like terms.
\$\$ \begin{aligned} (x \cdot 2x^2) + (x \cdot (-4x)) + (x \cdot 5) + (3 \cdot 2x^2) + (3 \cdot (-4x)) + (3 \cdot 5) \ = 2x^3 - 4x^2 + 5x + 6x^2 - 12x + 15 \ = 2x^3 + 2x^2 - 7x + 15 \end{aligned} \$\$
Multiplication Patterns
Two useful patterns for multiplying binomials without using FOIL or the combination method are the perfect square multiplying pattern and the difference of perfect squares multiplying pattern.
Simplifying $(x + 5)^2$ with FOIL becomes $x^2 + 5x + 5x + (5 \cdot 5) = x^2 + 10x + 25$. The coefficient of the $x$ in the solution is double the $+5$, whereas the constant in the solution is the square of $+5$. In general, $(x + a)^2 = x^2 + 2ax + a^2$.
Use the perfect square multiplying pattern to simplify $(x - 3)^2$.
Solution: The coefficient will be $2 \cdot (-3) = -6$, and the constant will be $(-3)^2 = +9$.
\$\$ x^2 - 6x + 9 \$\$
When the only difference between two binomials is that one has a $+$ between the two terms and the other has a $-$ between the two terms, there is a shortcut for multiplying the binomials.
Simplifying $(x - 5)(x + 5)$ with FOIL becomes $x^2 + 5x - 5x - 25 = x^2 - 25$. There is no $x$-term in the answer, and the
constant term is the negative square of the constant term of either of the binomials. In general, $(x - a)(x + a) = x^2 - a^2$.
Use the difference of perfect squares multiplying pattern to simplify:
\$\$ (x-3)(x+3) \$\$
Solution: Since the only difference between the two binomials is the sign between the two terms, this pattern can be used. The answer is $x^2 - 3^2 = x^2 - 9$.
Dividing Polynomials
Dividing polynomials requires a process very similar to long division for numbers.
\$\$ (2x^3 + x^2 - 11x + 12) \div (x + 3) \$\$
Step 1:
Set up for the long division process.
\$\$ (x+3)\overline{2x^3+x^2-11x+12} \$\$
Step 2:
Determine what you would need to multiply by the first term of the divisor ($x$ in this example) to get the first term in the dividend ($2x^3$ in this example). Since you would need to multiply $x$ by $2x^2$ to get $2x^3$, the first term of the solution is $2x^2$. Put that term over the $x^2$-term in the dividend.
\$\$ 2x^2 \$\$
\$\$ (x+3)\overline{2x^3+x^2-11x+12} \$\$
Step 3:
Multiply the $2x^2$ by the $x + 3$ to get $2x^3 + 6x.$ Put that product under the $2x^3 + x^2$. Subtract and bring down the $-11x$.
\$\$ \begin{array}{c} 2,\mathrm{x}^2\ \infty + 3\overline{\left|2,\mathrm{x}^3 + ,,\mathrm{x}^2 - 1,\mathrm{1x} + 1,\mathrm{2}\right|}\ -\left(2,\mathrm{x}^3 + 6,\mathrm{x}^2\right) \\ \hline \end{array} \$\$
\$\$ -5,\mathrm{x}^2 - 1,\mathrm{1x} \$\$
Step 4:
Determine what you would need to multiply by the first term of the divisor ($x$ in this example) to get the first term in the expression at the bottom ($-5x^2$ in this example). Since you would need to multiply $x$ by $-5x$ to get $-5x^2$, the second term of the solution is $-5x$. Put that term over the $x$-term in the dividend. Multiply the $-5x$ by
the $x + 3$, and put the product under the $-5x^2 - 11x$. Then subtract and bring down the $+12$.
Step 5:
Determine what you would need to multiply by the first term of the divisor ($x$ in this example) to get the first term in the expression at the bottom ($4x$ in this example). Since you would need to multiply $x$ by $4$ to get $4x$, the third term of the solution is $+4$. Put that term over the constant term in the dividend. Multiply the $+4$ by the $x + 3$, and put the product under the $4x + 2$. Then subtract. As there is nothing left to bring down, this final number at the bottom is the remainder. Since the remainder in this example is $0$, we sometimes say there is no remainder and that $x + 3$ divides evenly into $2x^3 - x^2 - 11x + 12$.
\$\$ \begin{array}{c|c} 2\mathbf{x}^2 - 5\mathbf{x} + 4\ \mathbf{x} + 3\overline{)2\mathbf{x}^3 + (-\mathbf{x}^2 - 11\mathbf{x} + 12} \ -(2\mathbf{x}^3 + 6\mathbf{x}^2) \ \hline \ -5\mathbf{x}^2 - 11\mathbf{x} \ -(-5\mathbf{x}^2 - 15\mathbf{x}) \ \hline \ \hline \ 4\mathbf{x} + 12 \ -(4\mathbf{x} + 12) \ \hline \ \hline \ \end{array} \$\$
Step 6:
You can check your answer by multiplying the solution by the divisor and then adding the remainder to see if the result is equal to the dividend.
\$\$ \begin{aligned} &(x+3)(2x^2 - 5x + 4) + 0 \ &= 2x^3 - 5x^2 + 4x + 6x^2 - 15x + 12 + 0 \ &= 2x^3 + x^2 - 11x + 12 \end{aligned} \$\$
What is the quotient and remainder (if any) of the following?
\$\$ (3x^3 + 8x^2 - 14x + 13) \div (x+4) \$\$
- (1) $3x^2 + 4x - 2$, remainder $5$ (2) $3x^2 - 4x + 2$, remainder $5$ (3) $3x^2 + 4x + 2$, remainder $6$ (4) $3x^2 - 4x - 2$, remainder $6$
Solution:
\$\$ \begin{array}{r} 3x^2 - 4x + 2 \ \hline ,, 3 ,, 4 \overline{, 3 ,, ^3 + ,, 8x^2 - 14x + 13} \ \hline -(3x^3 + 12x^2) \\ \hline \hline -4x^2 - 14x \ \hline -(-4x^2 - 16x) \ \hline \end{array} \$\$
\$\$ \begin{array}{r} 2x + 13 \ \hline -(2x + 8) \ \hline \end{array} \$\$
The answer is choice $(2)$.
Since Example $4$ is a multiple-choice question, an alternative way to do this one would be to check each of the answer choices. Multiply each potential solution by $x + 4$, and add the potential remainder. The answer choice that gives you $3x^3 + 8x^2 - 14x + 13$ is the correct one.
Checking choice $(2)$ would look like:
\$\$ \begin{aligned} &(x+4)(3x^2 - 4x + 2) + 5 \ &= 3x^3 - 4x^2 + 2x + 12x^2 - 16x + 8 + 5 \ &= 3x^3 + 8x^2 - 14x + 13 \end{aligned} \$\$
