Algebra: Understanding Polynomials
Polynomials are expressions with more than two variables, and the variables are usually unlike, having different powers/exponents and are combined by the four major operations: addition, multiplication, division and subtraction. For example 3x3 + 5x2 + 2x. If you know about quadratic expressions you’ll realize that a quadratic expression is pretty much a polynomial. However, before we get into the working out, there’s some background information that you should note:
You can’t add unlike terms, for example 3x2 + 2x ≠ 5x3. Even though both expressions have the variable x, you have to realize that x2 and x are unlike variables. You can’t subtract unlike terms either, thus 3x2 – 2x ≠ 1x. However, you can multiple and divide unlike terms. For example 3x2 × 2x = 6x3 and 3x2 ÷ 2x = 3/2 × x. You can only add and subtract like terms. For example 2x + 3x = 5x, 2x2 + 5x2 = 7x2.
Solve for ‘b’ in the following (x2 + bx − 2) (x + 3) = x3 + 6x2 + 7x – 6.
In order to solve for ‘b’ in the above question, we need to first consider the left hand side of the equation (it is easier to work with the left hand side). Then we expand it using the distributive law. Therefore:
(x2 + bx − 2) (x + 3) = (x2 × x) + (3 × x2) + (bx × x) + (3 × bx) + (-2 × x) + (-2 × 3)
Thus we should have: x3 + 3x2 + bx2 + 3bx – 2x – 6.
In this, since the ‘b’ is still unknown we can convert the following in a form that will make it easier or since there are like terms we can add and/or subtract them. For example bx2 and 3x2 are like terms, and in this case we can add the coefficients since we do not know what ‘b’ is it can be written as such (3 + b)x2. We apply the same concept for 3bx – 2x, which will be (3b – 2)x. Therefore we now have:
x3 + (3 + b)x2 + (3b – 2)x – 6 = x3 + 6x2 + 7x – 6.
Since the expression on the left hand side and the expression on the right hand side are equal, it means that the coefficients of each variable must also be equal. Sometimes this is called the concept of equating coefficients. This concept is useful for the topic partial fractions (which you don’t have to worry about unless you’re studying more advanced math). So, we note that:
3 + b = 6 or 3b – 2 = 7. You can use any one of these equations to solve for ‘b’.
3 + b = 6
3 – 3 + b = 6 – 3
0 + b = 3
b = 3
Sometimes the questions may come in the form of multiple choice but it doesn’t matter how it comes, once you understand the concept, then you’re good to go.
Questions may require you to rewrite expressions some times as difference of squares or deeper analysis of polynomials. The concept of difference of squares is quite easy. All you need to remember is this:
Difference of squares is defined as such: a2 – b2 = (a + b)(a – b).
Factorize the following:
25x2 – 4x2
So in order to work this, we need to remember the definition of difference of squares. We also need to identify our a2 and b2. We note from the question that 25x2 satisfied a2. This means there is a number or term that you can multiple by itself to produce 25x2. Thus, if we find the square root of the term 25x2, we get 5x which will be our ‘a’. Also the same principle applies for 4x2. Thus √4x2 = 2x and 2x will be our ‘b’. Therefore we note that (5x)2 – (2x)2 satisfies a2 – b2. Therefore we can rewrite the expression as such:
25x2 – 4x2 = (5x + 2x)(5x – 2x)
Sometimes the question may come in a way that requires you to think beyond just the concept of difference of squares to solve it.
16s4 − 4t2
In this example we only see a power of 2 on t but not on s. In this case what do you do? You need to try to rewrite the expression in a way that it satisfies a2 – b2. So looking at 16s4, we can rewrite it as such (4s2)2. Thus, it does not change the value of the expression and it now satisfied a2. We now have (4s2)2 – (2t)2 which satisfied a2 – b2. Therefore we can now use the concept of difference of squares.
16s4 − 4t2 = (4s2 + 2t)(4s2 – 2t).
Another way you will be tested in terms of polynomial is as such:
y5 − 2y4 − cxy + 6x
In the polynomial above, c is a constant. If the polynomial is divisible by y − 2, what is the value of c?
In order to solve the question. You have to look at the information that is given and ask yourself what you know about the information. They tell us that the polynomial is divisible by y – 2. That means it can be divided into the polynomial without leaving a remainder. With this, we check what is call the Factor Theorem (if you are unsure of this, please do further reading). This theorem states that if y – 2 can be divided into the polynomial without leaving a remainder then if we equate y – 2 to zero and solve, the result can be substituted in the polynomial and the results should be zero. Therefore: If x – a is a factor of f(x), then f(a) = 0. Let us apply it.
y – 2 = 0
y – 2 + 2 = 0 + 2
y + 0 = 2
y = 2
Therefore if 2 is substituted into y5 − 2y4 − cxy + 6x then it should be equal to zero. Thus:
(2)5 – 2(2)4 – cx(2) + 6x = 0
32 – 32 – 2cx + 6x = 0
-2cx + 6x = 0
-2cx = – 6x
Thus at this point in order to solve for c, we use the concept of equating coefficients. Therefore:
-2c = -6
-2c/-2 = -6/-2
Therefore c = 3
So hopefully, by now, you have a fairly good understanding of polynomials. As always, in math, the more you practice, the better you get at it. We’ll keep covering important topics in Algebra, so keep visiting our blog for more stuff!