Solve each polynomial inequality. Give the solution set in interv...

Question

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. x^5 + x^2 + 2 ≥ x^4 + x^3 + 2x

Accepted Answer

Hello, everyone. In this video, we're gonna be looking at this practice problem which states for the given inequality expression solve and write the solution set in interval notation. And the inequality is X to the fifth minus two X squared plus 24 greater than or equal to two X to the fourth minus X to the third plus 12 X, we're given four answer choices for this practice problem answer choice A is an interval from negative infinity to two with a bracket at two union with bracket 22 positive infinity. Answer choice B is bracket negative square root of three to positive square root of three bracket union with bracket two to positive infinity. Answer choice C is negative square root of three to positive infinity. And answer choice D is negative infinity to two to positive two with a bracket on the two. So in this case, to figure out which of these answer choices is correct. Let's take a look at our inequality which is X to the fifth -2, x squared plus greater than or equal to two X to the fourth minus X to the third plus 12 X So in this case, you can see that we have terms on both the left hand side and the right hand side. And in order to solve this inequality, I need to isolate X. So I'm gonna move all my terms to one side by subtracting by two X to the fourth, adding X to the third and subtracting 12 X from both sides. And when I move these terms over to the left hand side, I'm gonna write my equation in standard form so that I'm left with X to the fifth minus two, X to the fourth Plus X to the 3rd minus two, X squared minus 12 X plus Greater than or equal to zero. And from here, you can see that on my left hand side has X to the group to the degree of five, which means I need to start factoring out XS. So in this case, I'm gonna use the rational zero theorem which tells me that I can figure out which numbers are gonna be zero if I look at the factors of my constant, which in this case is 24 over the factors of my leading coefficient, which in this case, my leading coefficient is one. So that means that because the factors are over one, I really only care about the factors of my constant. So when we think of factors of 24 I have plus or -1 plus or minus plus or minus two plus or minus plus or minus three plus or minus eight plus or minus four plus or minus six. So in this case, you can see that we have a lot of factors. So I'm just gonna start with the smallest one, which in this case is one and see if it gives me a zero. So I have one to the fifth power minus two, multiplied by one to the fourth power plus one to the third power -2, multiplied by one squared -12 multiplied by one plus 24. So in this case, I have one -2, which is negative one plus one. So zero minus two brings me back down to negative two minus 12, negative 14 plus 24. So this equals 10. So I know that one is not E zero. So then I'm gonna try plugging in two. So that means I have two to the fifth power minus two, multiplied by two to the fourth power plus two to the third power minus two, multiplied by two squared minus 12, multiplied by two plus 24. And in this case, two to the fifth power is 32 minus two, multiplied by two to the fourth power which is 16 Plus 2 to the third, which is eight -2 multiplied by two squared, which is two multiplied by four, Which is 8 -12 multiplied by two, which is 24 of the plus 24. And in this case, two times 16 is 32. And from here, you can see that all the terms or all the numbers cancel out. So it leaves me with zero, which means that two is a Roop. So x equals two is a group, Which means I'm going to factor out X - and I'm gonna factor that out using the synthetic division. So I'll have a two and I'll have the coefficients of the expression. So I have one negative to positive one, -2, and positive 24. So I'll drop down the 12 times one is two. This is zero, zero times two is zero, is one, again, one times two is two to zero, two times zero is zero, negative 12 plus zero is negative 12 and two multiplied by negative 12 is negative 24. So this goes to zero. So that means that my expression is the X - that I divided out. And the resulting expression which is X to the 4th Plus zero, X to the 3rd Plus one, x squared plus zero, X minus 12. So if I get rid of the terms with a coefficient of zero, The expression is X to the 4th plus X squared minus 12. And now I just need to factor my new expression X to the fourth plus X squared minus 12. So when I factor it, I need two terms and the first terms of those factors need to multiply to X squared or X to the fourth. So if I do X squared multiplied by X square, that gives me X to the fourth. And my last terms need to multiply to negative 12. But add to the coefficient in front of the middle term, which is one. So I can try three and four. In this case, I need one of them to be negative. So I'll do negative three, multiplied by four. That gives me negative 12. And if I do negative three plus four, that gives me positive one. So I wanna say X squared minus three and X squared plus four on my factors. And I still have the X minus two. So you can see that from solving and factoring out the axes, I'm gonna get at least or at minimum at most um six solutions for or five solutions for X Because of the degree five. So from here, I'm gonna wanna solve for X Where I have X -2 is equal to zero X squared minus three is equal to zero and X squared plus four is equal to zero. So X minus two equal to zero. If I solve for X, I'm gonna add two, leaving me with X equals two, X squared minus three can add three to both sides. Leaving me with X squared is equal to positive three. If I take the square root of both sides, I have X is equal to plus or minus the square root of three. And this last one, X squared M plus four also tracked four from both sides. So I have X squared is equal to negative four. Well, I could take the square root, then I would go into imaginary numbers and I don't want imaginary values. So I'm gonna not solve that one and just ignore it because I have the square root of a negative number. So that leaves me with Negative Square Root of three, Politer Square Root of three or 2. And in this case, if we were solving an equation, these would be the solutions for X. But in this case, we have an inequality. So these become the numbers where we need to test intervals. So because we have three solutionss for X, we're gonna have four intervals. We have the first interval from negative infinity to negative square root of three. The second interval is negative square root of three to the positive square root of three. And then we have from the square root of 3 to positive two And the final interval is from 2 to Infinity. So we need to test values in each of these intervals and see if our inequality is true. So starting with the first interval, negative infinity to negative square root of three, we can try the value X at -4. So we would have negative 4 - multiplied by negative four squared minus three multiplied by negative four squared plus four. So negative four minus two is negative six, multiplied by negative four squared is positive 16 minus three is 13, negative four squared is again positive 16 plus four is 20. And recall that we want this to be greater than or equal to zero. Well, in this case, we have a negative and no other negative to cancel it out. Which means that this will give us a negative answer which is not greater than or equal to zero. So this interval does not fulfill the inequality. If we look at the next interval, negative square root of 32, positive square root of three, you can try the value at X equals.

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. x^5 + x^2 + 2 ≥ x^4 + x^3 + 2x

Key Concepts

Polynomial Inequalities

Interval Notation

Sign Analysis

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