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## Algebra II (2018 edition)

# Finding zeros of polynomials (2 of 2)

CCSS.Math: , , ,

Sal uses an alternative method to find the zeros of p(x)=x⁵+9x³-2x³-18x=0.

## Want to join the conversation?

- at around2:00how does he know it will be x to the fourth and x squared?(8 votes)
- When you multiply (x^2 + a) with (x^2 + b) , you get

x^2(x^2 + b) + a(x^2 + b) using the distributive property.

x^2(x^2 +b) = x^(2*2) + bx^2

a(x^2 +b) = ax^2 + ab

Adding these two will give us the value of (x^2 +a)(x^2 +b).

x^(2*2) +bx^2 + ax^2 + ab = (x^2 +a)(x^2 +b)

x^4 + x^2(a+b) + ab = ( x^2 +a)(x^2 +b)(10 votes)

- or we could've just added the cubed terms at the beginning?(7 votes)
- You could have... it doesn't save you any effort. You still need to do the factoring to get to the factors Sal has found.(7 votes)

- My teacher wants us to find the zeroes of P(x) where P(x)=x^4-3x^2-14x-12

How would you solve this if you can't factor it by grouping?(10 votes) - why isnt x^2=-9 an answer?(3 votes)
- x^2 = -9 does yield 2 roots for this polynomial, neither of which are real. See the previous video where all 5 roots are discussed.(4 votes)

- how would i find the zeros for a polynomial without a constant factor? For example, 5x-x^3.(2 votes)
- I'm going to assume this is really 5x-x^3 = 0

You need an equation to be finding the roots (the x-intercepts).

factor out an x: x (5 - x^2) = 0

x = 0 is one root

5-x^2 = 0

Add x^2: 5 = x^2

take square root of both sides: +/- sqrt(5) = x

Other 2 roots are: x = +/- sqrt(5)(3 votes)

- How do I revise what I've learnt here?(3 votes)
- I believe there are two exercises a video after this. You can revise by aiming for 100% on both exercises to make sure you've learned it. Hope I have helped :)(0 votes)

- For the equation f(x)= x^3-5x^2-25x+125, I found only 2 of 3 real solutions. the solutions are x=5 and x=-5. My graphing software also shows x=5 and x=-5 as my x-intercepts on the graph of the equation. Are there is a complex solution for the equation or I have to guess the last real solution for the equation? I am being taught that there 3 possible real solutions or 1 real solution and 2 complex solutions for degree 3 polynomial functions.(1 vote)
- x = -5 is a single root, while x = 5 is a double root.

x^3 - 5x^2 -25x + 125 = 0

( x^3 + 125 ) + ( -5x^2 - 25x ) = 0

Factor the first binomial as the sum of cubes.

Factor the common factor of -5x from the second binomial.

Then factor out the GCF of ( x + 5 ) and also factor the trinomial that remains to get

( x + 5) ( x - 5 )^2, which indicates the 3 real roots.

Hope this helps!(3 votes)

- At2:35he said that there is a difference of squares, and you can further factorize it. Why can't he also factorize the (x^2+9) if he really wanted to fully factorize the result? Why only the "difference of squares"?(1 vote)
- x^2 + 9 is a Sum of 2 Squares. This is not factorable.

To try and factor it...

Start by inserting a middle term of 0x. This doesn't change the binomial, it just makes it clearer as to what we need to try and create.

x^2 + 0x + 9

We need to find 2 numbers that multiply and = 9, but also add to 0.

The possible factors of 9 are: 1 and 9; or 3 and 3

Neither pair will add to 0.

Hope this helps.(1 vote)

- hi! i seem to have found a polynomial where the rational root theorem or any other form of factoring cannot work. the problem is y=2x^3-5x^2+10x-3. i'm hoping sal will notice this comment and help a kid out(1 vote)
- There are plenty of polynomials with integer coefficients and only irrational or nonreal roots. This is one of them. The roots are listed here:

https://www.wolframalpha.com/input/?i=roots+of+2x%5E3-5x%5E2%2B10x-3(1 vote)

- I have a problem it says y=3x^3+5x^2-107x+35, it wants me to find all the zeros in the function. How would I solve this??(1 vote)

## Video transcript

- [Voiceover] In the last video,
we factored this polynomial in order to find the real roots. We factored it by grouping, which essentially doing
the distributive property in reverse twice. And I mentioned that there's
two ways you could do it. You could actually, from the get go, add these two middle degree terms, and then, think about it from there. So, what I thought I'd
do is just a quick video on that alternative. So, if we add, instead of grouping, if we add these middle two terms. Actually, I'll just focus on the fourth degree polynomial here. We know that we have an x out front. This fourth degree polynomial
is going to simplify to x to the fourth plus seven x squared minus 18. If we wanna factor this, we
could recognize a pattern here. You probably remember. Hopefully you remember. If you don't, then you might wanna review your factoring polynomials. But if you have x plus a times x plus b, that's going to be equal to, x is going to be equal to x squared plus the sum of those two numbers, a and b, as being the coefficient of the x term plus the product of those two numbers. If you just multiply this out, this is what you would get. But if this was x squared
plus a times x squared plus b, instead of this being x squared, this would be x to the fourth. Instead of this being x,
this would be x squared, which is exactly the pattern we have here. So, what two a's and b's
that if I add them up, I would get seven, and if I were to take their product, I get negative 18? Well, since their product
is negative, we know that they are of different signs. One will be positive,
one will be negative. And since their sum is positive, we know that the larger of the two numbers is going to be positive. So, what jumps out at me is nine times negative two. You multiply those, you get negative 18. You take their sum, you get seven. So, we can rewrite this, just
looking at this pattern here as x squared plus nine times x squared minus two. I could say plus negative two. That's the same thing
as x squared minus two. And then, that's exactly what we got right over here. Of course, you have this x out front that I didn't consider right over here. And then, this, as we did
in the previous video, you could recognize as
a difference of squares and then, factor it further
to actually find the roots. But I just wanted to show
that you could solve this by regrouping, or you can solve this by, I guess you could say, more
traditional factoring means. And notice this nine and negative two, this is what was already broken up for us, so we could factor by regrouping.