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# Zeros of polynomials (with factoring): common factor

When a polynomial is given in factored form, we can quickly find its zeros. When it's 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 by first taking a common factor and then using the sum-product pattern.

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• How did we get (x+3)(x-2) from (x^2+x-6)? I can see where the +3 and -2 came from, but what's going on with the x^2+x part? And, how would I apply this to an equation such as (x^2+7x-6)?
• There are numerous ways to factor, this video covers getting a common factor. This will not work for x^2 + 7x - 6. Ic an tell you a way that works for it though, in fact my prefered way works for all quadratics, and that i why it is my preferred way.

Anyway, with x^2 + x - 6 Sal used this method.

if you have a quadratic ax^2 + bx + c where a, b and c are some numbers, you want to find two more numbers d and e. you want d + e = b and d * e = a * c, so it's kinda a guess and check to find them. Anyway, you then change the original from ax^2 + bx + c to ax^2 + dx + ex + c, so you replace bx with dx + ex

In x^2 + x - 6 a = 1, b = 1 and c = -6. With this we wand d + e = 1 and d * e = 1 * -6 = -6. You usually want to start with the multiplication one, so what are all the numbers that multiply to equal equal -6?

1 * -6
-1 * 6
2 * -3
-2 * 3
3 * -2
-3 * 2

Now of these, do any work out to add together to be 1? Yes, -2 and 3 do. So you change the equation.

x^2 + x - 6
x^2 - 2x + 3x - 6

You could make -2x + 3x into 3x - 2x if you want, it works either way. I encourage you to try it the other way after I show you this way

Anyway, you ant to put parenthesis around the left two numbers and the right two numbers.

x^2 - 2x + 3x - 6
(x^2 - 2x) + (3x - 6)

Keep in mind x^2 + 3 - 2x - 6 would become (x^2 + 3) + (- 2x - 6) i mention it because the - could throw you off. Anyway, continuing

(x^2 - 2x) + (3x - 6)

From (x^2 - 2x) you can pull out a x to get x(x - 2) and a 3 from the other to get 3(x - 2) so now you have

x(x-2) + 3(x - 2)

Notice how both have x-2? That was what our d and e did, so if you can find a d and e this will always work out. Sometimes you won't be able to find d and e though. After this you can change it to

x(x - 2) + 3(x - 2)
(x + 3)(x - 2)

If you don't understand how to do that I can explain, as well as anything else you don't understand. Anyway, that's how to get from x^2 + x - 6 to (x+3)(x-2)
• What if you have a function that = x^3 + 8 when finding the zeros? Would you just cube root?
• You probably could if you had a function that was a cube root. Otherwise, if the function can't be factored, you write the answer out in cube root form
• Could you also factor 5x(x^2 + x - 6) as 5x(x+2)(x-3) = 0 to get x=0, x= -2, and x=3 instead of factoring it as 5x(x+3)(x-2)=0 to get x=0, x= -3, and x=2?
• No because -3 and 2 adds up to -1 instead of 1. They have to add up as the coefficient of the second term.
• We want to find the zeros of this polynomial: p(x)=2x3+5x2−2x−5 Plot all the zeros (x-intercepts) of the polynomial in the interactive graph.
• The first way to approach this is to see if you can factor out something in first two terms and second two terms and get another common factor. So p(x)= x^2 (2x + 5) - 1 (2x+5) works well, then factoring out common factor and setting p(x)=0 gives (x^2-1)(2x+5)=0. From there, note first is difference of perfect squares and can be factored, then you use zero product rule to find the three x intercepts.
• how did you get -6 out of -30
• You can divide it by 5
• I have almost this same problem but it is 5x² -5x -30. What should I do there? Should I group them together?
• When you are factoring a number, the first step tends to be to factor out any common factors, if possible. In this problem that common factor is 5, so we can factor it out to get 5(x² - x - 6). Then we can factor again to get 5((x - 3)(x + 2)). This isn't the only way to do this, but it is the first one that came to mind.
• how to find more values of x
• In this example, he used p(x)=(5x^3+5x^2-30x)=0. Using that equation will show us all the places that touches the x-axis when y=0. But if we want to find all the x-value for when y=4 or other real numbers we could use p(x)=(5x^3+5x^2-30x)=4. Simply replace the f(x)=0 with f(x)= ANY REAL NUMBER. I hope this helps.
• p(x)=2x^(3)-x^(2)-8x+4

is there another way to factor polynomials with 4 terms

without seperating into 2 group , first group : first term, second term ,

second group : third term and fourth term.
• There might be other ways, but separating into 2 groups is useful for 90% of the time.
• the only factors of the function f are (x+3)^2 and (3x-2) what is the product of its three zeroes?
• (𝑥 + 3)² and (3𝑥 − 2) are the only factors of 𝑓.

Thus, 𝑓(𝑥) = (𝑥 + 3)²⋅(3𝑥 − 2)

In practice, 𝑓 only has two zeros.
𝑓(𝑥) = 0 ⇒ 𝑥 = −3 or 𝑥 = 2∕3

But, 𝑥 = −3 is what we call a double zero (because the factor (𝑥 + 3) occurs twice in the factorization of 𝑓).

So technically, 𝑓 has the three zeros
𝑥₁ = −3
𝑥₂ = −3
𝑥₃ = 2∕3

And their product is (−3)⋅(−3)⋅2∕3 = 6