For each polynomial function, find all zeros and their multipl...

Question

For each polynomial function, find all zeros and their multiplicities. $ƒ (x) = (x^{2} + x - 2)^{5} (x - 1 + \sqrt{3})^{2}$

Accepted Answer

Hello, everyone. We are asked to identify the zeros and their multiplicities. For the following function F of X equals open parentheses. X squared plus two, X minus 24 closed parentheses raised to the seventh power multiplied by open parentheses, X minus two plus radical five, closed parentheses raised to the fourth power. Our answer choices are a negative two plus radical five with a multiplicity of four B zero with a multiplicity of seven two plus radical five with a multiplicity of four C negative 46 with a multiplicity of seven, two minus radical five with a multiplicity of four D negative 64 with a multiplicity of seven, two minus radical five with a multiplicity of four. First, I'm going to rewrite my function. So F of X equals open parentheses, X squared plus two, X minus 24. Close parentheses, raise the seventh power multiplied by open parentheses. X minus two plus radical five closed parentheses raised to the fourth power. The exponents on our parentheses will tell us the multiplicity of the answers that come from those parentheses. So the solutions from our first set of parentheses will have a multiplicity of seven and the solutions from the second set of parentheses will have a multiplicity of four to find what the solutions are. We will set each factor equal to zero as we did way back when, when we started solving for X. So first, I'm going to do X squared plus two X minus equals zero. This is a trinomial and it looks factor. So I will be factoring it. So I'm gonna open two sets of parentheses and set them equal to zero. I recall when I factor my first term in my binomials have to multiply to my first term in the trinomial. So X multiplied by X will give me X squared. Then the second term in the binomials must multiply to, in this case negative 24 but add to a positive two, 24 has a lot of factors. But the ones that came to mind are a positive six, the negative four, those will multiply the negative 24 add to a positive two. Setting each factor here equal to zero, X plus six equals zero and X minus four equals zero. Solving for X, I first subtract six from both sides from X plus six equals zero. I get that X equals negative six for X minus four equals zero. I add four to both sides get that X equals four. So the first part of our solution is X equals negative six, X equals four. And this is gonna have a multiplicity of seven because it comes from the parentheses that had the seven. This is part of our solution. We're not done yet. Next, I need to set X minus two plus radical five equal to zero to find out what the value is with a multiplicity of four. There's a couple ways I can move this. So X ends up by itself, but I'm gonna do it one term at a time. So first, I'm gonna add two to both sides. So this gives me X plus radical five equals two. Next, I'm subtracting radical five from both sides. So I get that X equals two minus radical five. So this is our other X value. So this solution because it comes from the parentheses with the exponent of four has a multiplicity of four. So now we'll see which ones of these match our answer choices. So we have negative six and four with a multiplicity of seven, then two minus radical five with a multiplicity of four. And that is answer choice. D have a nice day.

For each polynomial function, find all zeros and their multiplicities. $ƒ (x) = (x^{2} + x - 2)^{5} (x - 1 + \sqrt{3})^{2}$

Key Concepts

Polynomial Zeros

Multiplicity of Zeros

Factoring and Solving Quadratic Expressions

For each polynomial function, find all zeros and their multiplicities. ƒ(x)=(x2+x−2)5(x−1+3)2ƒ(x)=(x^2+x-2)^5(x-1+\sqrt3)^2

Key Concepts

Polynomial Zeros

Multiplicity of Zeros

Factoring and Solving Quadratic Expressions

For each polynomial function, find all zeros and their multiplicities. $ƒ (x) = (x^{2} + x - 2)^{5} (x - 1 + \sqrt{3})^{2}$