A polynomial is a mathematical expression with many terms ("poly" means "many" and "nomial" means "names" or "terms").
A polynomial can have any combination of operators (addition, subtraction, multiplication or division), constants, variables and exponents, but never division by a variable. Remember, this means that expressions with negative exponents can never be polynomials because $a^{-x}$a−x is the same as $\frac{1}{a^x}$1ax.
Polynomials are usually written in descending order, starting with the term with the greatest power and ending with the term with the least (or no) power. For example, in the polynomial $8x+4x^8-2x^2+4$8x+4x8−2x2+4, the powers are all jumbled. To put it in order, we would rewrite it as $4x^8-2x^2+8x+4$4x8−2x2+8x+4.
Examples of expressions that ARE polynomials | Examples of expressions that ARE NOT polynomials |
---|---|
$5x^2+\frac{4}{3}x-7$5x2+43x−7 | $\frac{4}{x-3}$4x−3 |
$-18$−18 | $3+\frac{1}{x}$3+1x |
$3x$3x | $4x^3-\frac{1}{x^7}+8$4x3−1x7+8 |
$4c-8cd+2$4c−8cd+2 | $\frac{7}{8}x^{-2}+5$78x−2+5 |
$22x^6+12y^8$22x6+12y8 | $\sqrt{x}$√x |
$7g+\sqrt{12}$7g+√12 | $12f^3g^{-4}\times h^6$12f3g−4×h6 |
If an expression contains terms that divide by a variable (i.e. an algebraic term), they are NOT polynomials.
Degree: The largest exponent or the largest sum of exponents of a term within a polynomial. For example, in the polynomial $x^3+4x^2-9$x3+4x2−9, the greatest power of $x$x is $3$3, so the degree in this polynomial is $3$3.
Leading coefficient: The coefficient of the first term of a polynomial written in descending order of exponents. For example, in $5x-7$5x−7, the leading coefficient is $5$5 and in $-x^5-2x^4+4$−x5−2x4+4, the leading coefficient is $-1$−1.
Constant term: the term in a polynomial that has no variables (degree of 0). For example, in the polynomial $4y^8+2xy-4x-\frac{2}{3}$4y8+2xy−4x−23, the constant term is $-\frac{2}{3}$−23.
Is $2x^3-4x^5+3$2x3−4x5+3 a polynomial?
Yes, it is a polynomial.
No, it is not a polynomial.
For the polynomial $P\left(x\right)=\frac{x^7}{5}+\frac{x^6}{6}+5$P(x)=x75+x66+5
The degree of the polynomial is: $\editable{}$
The leading coefficient of the polynomial is: $\editable{}$
The constant term of the polynomial is: $\editable{}$
Adding and subtracting polynomials is very similar to the process of simplifying algebraic expressions by collecting like terms.
Just like with any algebraic expression, we can only add and subtract like terms.
Let's run through the process by looking at an example. Let's say we want to find the difference between two polynomials: $P\left(x\right)=7x^3+4x^2-4$P(x)=7x3+4x2−4 and $Q\left(x\right)=7x^3+8x^2-2x-8$Q(x)=7x3+8x2−2x−8.
1. Start by writing out the equation we want to simplify:
$P\left(x\right)-Q\left(x\right)=7x^3+4x^2-4-\left(7x^3+8x^2-2x-8\right)$P(x)−Q(x)=7x3+4x2−4−(7x3+8x2−2x−8)
2. Collect the like terms, taking any negative symbols into account.
Don't forget to subtract the entire polynomial. We must write the polynomial being subtracted in parentheses to remind us that the negative applies to every term being subtracted.
Remember if there is a term with no corresponding term in the other polynomial, we can treat this as a value of zero. For example, $P\left(x\right)$P(x) does not have a term with $x$x but $Q\left(x\right)$Q(x) does.
We can present it vertically in a table and align like terms:
$7x^3$7x3 | $+$+ | $4x^2$4x2 | $+$+ | $0$0 | $-$− | $4$4 | |||
$-$− |
$($( |
$7x^3$7x3 | $+$+ | $8x^2$8x2 | $-$− | $2x$2x | $-$− | $8$8 |
$)$) |
$0$0 | $-$− | $4x^2$4x2 | $+$+ | $2x$2x | $+$+ | $4$4 |
or we can simply distribute the negative and group the like terms:
$7x^3+4x^2-4-\left(7x^3+8x^2-2x-8\right)$7x3+4x2−4−(7x3+8x2−2x−8) | $=$= | $7x^3+4x^2-4-7x^3-8x^2+2x+8$7x3+4x2−4−7x3−8x2+2x+8 |
$=$= | $7x^3-7x^3+4x^2-8x^2+2x-4+8$7x3−7x3+4x2−8x2+2x−4+8 | |
$=$= | $-4x^2+2x-4$−4x2+2x−4 |
You can choose what method you like.
3. Write out the solution
$P\left(x\right)-Q\left(x\right)=-4x^2+2x+4$P(x)−Q(x)=−4x2+2x+4
Simplify $\left(3x^3-9x^2-8x-7\right)+\left(-7x^3-9x\right)$(3x3−9x2−8x−7)+(−7x3−9x).
If a picture frame has a length of $8x^2-9x+3$8x2−9x+3 and a width of $6x^3+9x^2$6x3+9x2, form a fully simplified expression for the perimeter of the rectangular picture frame.
If $P\left(x\right)=3x^2+7x-6$P(x)=3x2+7x−6 and $Q\left(x\right)=6x-7$Q(x)=6x−7, form a simplified expression for $P\left(x\right)-Q\left(x\right)$P(x)−Q(x).
Recall that to distribute an expression like $2\left(x-3\right)$2(x−3)we use the distributive property: $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC
Now we want to look at how to multiply two binomials together, such as $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d).
Let's look at $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) for example and see how this distribution works visually before we look at the algebraic approach. We will use the area of a rectangle. We can consider the expression $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) to represent the area of a rectangle with side lengths of $x+5$x+5 and $x+2$x+2 as seen below.
Another way to express the area would be to split the large rectangle into four smaller rectangles. Since the area of the whole rectangle is area of the sum of its parts we get the following:
$\left(x+5\right)\left(x+2\right)$(x+5)(x+2) | $=$= | $x^2+2x+5x+10$x2+2x+5x+10 |
$=$= | $x^2+7x+10$x2+7x+10 |
We don't want to have to draw a rectangle every time, so below we'll look at the algebraic approach.
When we multiply binomials of the form $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) we can treat the second binomial $\left(cx+d\right)$(cx+d) as a constant term and apply the distributive property in the form $\left(B+C\right)\left(A\right)=BA+CA$(B+C)(A)=BA+CA. The picture below shows this in action:
As you can see in the picture, we end up with two expressions of the form $A\left(B+C\right)$A(B+C). We can distribute these using the distributive property again to arrive at the final answer:
$\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) | $=$= | $ax\left(cx+d\right)+b\left(cx+d\right)$ax(cx+d)+b(cx+d) |
$=$= | $acx^2+adx+bcx+bd$acx2+adx+bcx+bd | |
$=$= | $acx^2+\left(ad+bc\right)x+bd$acx2+(ad+bc)x+bd |
We can actually be even more efficient in our algebraic approach by taking a look at the step below.
$\left(ax+b\right)\left(cx+d\right)=acx^2+adx+bcx+bd$(ax+b)(cx+d)=acx2+adx+bcx+bd
Notice that we have multiplied every term in the first bracket by every term in the second bracket. In general, that is what is required to multiply two polynomials together. We often use arrows as shown below to help us get every pair.
By distributing in this way we will get the result $x^2+2x+5x+10=x^2+7x+10$x2+2x+5x+10=x2+7x+10, the same result we would have obtained using the previous method. You may prefer to use this alternate method since it is more efficient.
Distribute and simplify $\left(x-3\right)\left(x+4\right)$(x−3)(x+4) .
Think: We need to multiply both terms inside $\left(x-3\right)$(x−3) by both terms inside $\left(x+4\right)$(x+4).
Do:
$\left(x-3\right)\left(x+4\right)$(x−3)(x+4) | $=$= | $x\left(x+4\right)-3\left(x+4\right)$x(x+4)−3(x+4) | |
$=$= | $x^2+4x-3x-12$x2+4x−3x−12 |
We can jump right to this step using the short-cut mentioned above |
|
$=$= | $x^2+x-12$x2+x−12 |
Distribute and simplify the following:
$\left(x+2\right)\left(x+5\right)$(x+2)(x+5)
Distribute and simplify the following:
$\left(7w+5\right)\left(5w+2\right)$(7w+5)(5w+2)
Distribute and simplify the following:
$\left(2n+5\right)\left(5n+2\right)-4$(2n+5)(5n+2)−4
What we know about multiplying two binomials, can be extended to multiply and two polynomials.
Every term in one pair of parentheses has to be multiplied by every other term in the other pair of parentheses.
Distribute and simplify $\left(x^3-4\right)\left(x^2+3x+5\right)$(x3−4)(x2+3x+5). What do you notice about the degree of the new polynomial?
Think: We need to multiply every pair of terms between the two parentheses. Then we will distribute and simplify.
Do:
$\left(x^3-4\right)\left(x^2+3x+5\right)$(x3−4)(x2+3x+5) | $=$= | $x^3\left(x^2+3x+5\right)-4\left(x^2+3x+5\right)$x3(x2+3x+5)−4(x2+3x+5) |
$=$= | $x^3\times x^2+x^3\times3x+x^3\times5-4x^2-4\times3x-4\times5$x3×x2+x3×3x+x3×5−4x2−4×3x−4×5 | |
$=$= | $x^5+3x^4+5x^3-4x^2-12x-20$x5+3x4+5x3−4x2−12x−20 |
Reflect: When we multiplied a quadratic (degree $2$2) and a cubic (degree $3$3) we ended up with a polynomial with degree$5$5. In general, leading terms of the product will be the product of the leading terms. This means using our laws of exponents, the degree will be the sum of the degrees of the two polynomials.
In general, we can say that
degree $\left(P(x)\times Q(x)\right)=$(P(x)×Q(x))= degree$\left(P(x)\right)$(P(x)) + degree$\left(Q(x)\right)$(Q(x)),
where $P(x)$P(x) and $Q(x)$Q(x) are polynomials.
Distribute $\left(a+2\right)\left(5a^2-2a+2\right)$(a+2)(5a2−2a+2).
Consider $R\left(x\right)$R(x), the product of the polynomials $P\left(x\right)=3x^5-3$P(x)=3x5−3 and $Q\left(x\right)=-2x^7+5x^5+6$Q(x)=−2x7+5x5+6.
What is the degree of $R\left(x\right)$R(x)?
What is the constant term of $R\left(x\right)$R(x)?
Is $R\left(x\right)$R(x) a polynomial?
Yes
No