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# Finding zeros of polynomials (2 of 2)

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

## Want to join the conversation?

• at around how does he know it will be x to the fourth and x squared?
• 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)
• 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?
• or we could've just added the cubed terms at the beginning?
• 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.
• why isnt x^2=-9 an answer?
• 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.
• how would i find the zeros for a polynomial without a constant factor? For example, 5x-x^3.
• 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
take square root of both sides: +/- sqrt(5) = x
Other 2 roots are: x = +/- sqrt(5)
• How do I revise what I've learnt here?
• 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 :)
• 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!
• At he 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)