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### Course: Class 10 (Old)>Unit 2

Lesson 3: Division algorithm for polynomials

# Zeros of polynomials (with factoring): grouping

When a polynomial is given in factored form, we can quickly find its zeros. When its given in expanded form, we can factor it, and then find the zeros! Here is an example of a 3rd degree polynomial we can factor using the method of grouping.

## Want to join the conversation?

• How do we know which do numbers are suppose to be grouped together?
• one thing you could do is order the numbers by degree and then group them together from left to right
• what is the derivative of x
• The derivative is the slope of the function at a certain position. If the function is at a straight line, the slope will never change. This means that the derivative is a constant. In the save of f(x) = x, the derivative is one. At any x value, the slope of the graph will be one.
• What would the case be for (x^2-4) ?
• You should eventually learn about the difference of squares, which basically says if you have something like (x^2 - a) you could factor it into (x - sqrt(a)(x - sqrt(a)) so this is really useful if a is a perfect square. It is here in this case.

So (x^2 - 4) = (x-2)(x+2)

Does that help? If you want to know why the rule works just look at what happens if you FOIL (x-a)(x+a)

EDIT

Also works if a number is in front of the x so (ax+b)(ax-b) = (ax)^2 - b^2
• Do you yeet grouping out the math window?
• I think it's in the next video.
(1 vote)
• I only factored to (x+1)(x^2-9) and then did x+1=0, x=-1; and x^2+9=0, x= +/- 3 (i.e. x=-3 or 3). This led to the same answers even though it wasn't completely factored. Is that bad?
• That's fine. Even if you factor it fully, you would've gotten the same answer.
• is (x-3)(x+3) the same thing as (x+3)(x-3)?
• Yes, multiplication is commutative, so a*b = b*a. (a and b can be numbers or functions.)
• does anybody know why we need to factor x^2-9 completely, cant u just call an x = 3 and 3^2 - 9 = 0? I supposed u need to find the roots but what really are these roots and why does factoring these to something like (x+b) even become a root?
• You generally start with a quadratic equation f(x)=x^2-9. This forms a parabola. Without a middle term, we know that the y intercept is the vertex, so setting x=0 we know that the vertex (and y intercept) is (0,-9). If we want to find the x intercepts (zeros, roots, or solutions) we set y=0, so we end up with x^2-9=0. Since this is a difference of perfect squares, we get (x+3)(x-3)=0. Using the 0 product rule, we know either x+3=0 or x=-3 or x-3=0 or x=3. We also note that (-3)^2-9=0 as well as 3^2-9=0.
• Does grouping mean fragmenting an expression(numerical or algebraic) into the products of two factors? OR something else?
• No, you're referring to factoring. Grouping is a trick that helps with factoring, it is not factoring itself.

Say I have x³+x²-2x-2. No two of these terms have a common factor. However, x³ and x² do have a common factor (of x²) as do -2x and -2 (of -2). Grouping refers to factoring only these sub-expressions, like this
x²(x+1)-2(x+1).

We can then treat this expression as having only two terms with a common factor of x+1, and factor out the x+1 as
(x+1)(x²-2).
• I read that an irreducible polynomial of the form (x^n) + (x^(n-1))+(x^(n-2))...+x^2+x+1(notice all the coefficients are one) where n can be written as 2^k where k is an integer "can be expressed using a series of quadratic equations. Not by representing it as a product of of simpler factors, because that's not possible[as it is irreducible] but by using equations whose coefficients solve other equations." I cannot make sense of this statement, even though I have tried several times(maybe the quadratics have roots that correspond to this equation, but this seems wrong, as a quartic, for example may have a fourth root, which is impossible to get in a quadratic equation). This statement is absolutely key to what I am reading, so I would be extremely grateful for any help. I'm sorry if I made this question difficult to read, I am trying to avoid run-on sentences but it's hard! Thanks in advance!
• "Can be expressed using a series of quadratic equations" is very ambiguous, since it doesn't say what operations are allowed between the quadratics. I can only guess whether they mean a composition of quadratics, or a polynomial in a quadratic, or something else.

However, it's worth noting that there are very few polynomials like this. The polynomial which is just a sum of powers like this is irreducible only if the degree is one less than a prime. Since you insist that the degree looks like 2^k, this means we're looking for primes of the form (2^k)+1, or Fermat primes. And there are only five known Fermat primes: 3, 5, 17, 257, and 65537. It is suspected that there are no more, but this is still an open problem in math.

It would make some sense if your text meant that we are to compose a sequence of quadratic functions, since composing a bunch of quadratics must yield a polynomial whose degree is a power of 2. But I'm not aware of anything useful that could be gained by writing the polynomial like this, and it seems like it is possible for polynomials whose degree is any power of 2, not just the irreducible ones. Also, it seems like the coefficients are just rational numbers, not anything more complex.

My only other guess is that it means you can write the polynomial as a product of quadratics whose roots are complex conjugate pairs. These quadratics would indeed have 'coefficients that solve other equations', but this is possible for any polynomial of even degree. It's not special for polynomials of your form, or with degree of a power of 2.

I will say to take heart; it's not you. Unless there is a lot of clarifying context that you left out, this statement is almost meaningless as it is written, and it's all the worse if the statement is indeed 'absolutely key'. I would reject the text you're reading and find a better one.