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

Question

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. x⁴ + 6x² + 1 ≥ 4x³ + 4x

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 fourth plus 54 X squared plus 81 greater than or equal to 12 X to the third plus 108 X. And for this problem, we're given four answer choices. Answer choice A is an interval from negative infinity to positive infinity. Answer choice B is another interval from negative infinity to three union with three to positive infinity. Answer choice C is the number three and answer choice D is no solution. So in order to figure out which of these answer choices is correct, let's take a look at our inequality which is X to the 4th Plus 54 x squared plus 81 greater than or equal to 12 X to third plus 108 X. So you can see that in this inequality, in order to find the solution, we need to isolate X. But right now we have XS on both the left hand side and the right hand side. So I'm going to move the terms on the right hand side over to the left hand side by subtracting by 12 x To the 3rd and 108 x from both sides. And when I do that subtraction, I'm gonna write my equation in standard form. So I'm gonna have the degree start with the highest and then descending. So I'll have X to the fourth minus 12 X to the third plus 54 X squared -108 x plus greater than or equal to zero. And now you can see that my inequality has all the terms on the left hand side and X has a degree of four. So that means that I need to start factoring out X's so that I can isolate X. So I'm going to begin isolating X or factoring out X using synthetic division. And I'm gonna figure out which number to divide by using the Rational zero Theorem. So the rational zero Theorem tells me a set of numbers that are probably gonna make my equation equal zero. And it's the factors of my constant term which in this case is 81 over the factors of my leading coefficient, which in this case is one. So that means I really only care about the factors of 81. So the factors of 81 are plus or minus one plus or minus 81 plus or minus nine plus or minus three or plus. Or minus 27. So in this case, I'm going to start by trying three. So I'm saying X equals three is a zero, which means that I'm gonna be factoring out X minus three. But let's make sure that three is an actual zero by plucking in three into my equation. So I'll have three to the fourth minus 12, multiplied by three to the third plus 54 multiplied by three squared minus 108 multiplied by three plus 81 Greater than are we no longer need the greater than, or equal to? Because I just wanna make sure it equals zero 3 to the fourth is 81 -12 multiplied by three to the third, which is 27 Plus 54, multiplied by nine -108 multiplied by three, which is 324. We still have the plus 81. So I have 81 minus 12 multiplied by 27 which is 324 plus 54 multiplied by nine, which is 486 minus plus 81. And once I add 81 minus 324 plus 486 minus 324 plus 81 I get zero. So I know that I can divide out X minus three. So I'll do that using synthetic division. So I'll do write a three in this box and then pull the coefficients From my expression. So the coefficients are 1 -12, positive 54, -108 And positive 81. So once I pull three, I can't divide one by three. So that one just gets carried down. And now I have three multiplied by one is three, negative 12 plus three is negative +93, multiplied by negative nine is negative 27 and 54 27 is positive 27, multiplied by positive 27 is 81 and negative 108 plus 81 is negative three, multiplied by negative 27 Is -81. So this becomes zero. So that means that my resulting expression is the X -3, then I multiplied out with what was left over Which was X to the 3rd -9 x squared Plus 27 x -27. And that means I need to simplify this blue expression or X to the third minus nine X squared plus 27 X minus 27. But if you look at it, we can use the cube of differences to simplify it. So the cube of differences follows the form that if we have an equation A to the third minus three A squared multiplied by B plus three A B squared minus B to the third. It can be rewritten as A minus B inside parentheses to the third power. So in our case, A would be X and B would be three. So we can test this by looking at our terms. So you can see that the first term is A to the third, which is X to the third. Our second term is negative three multiplied by A squared multiplied by B. So B is three. So we have negative three times multiplied by three is negative nine and A squared is X squared. And we do have negative nine X squared. The third term is three multiplied by a A is X and B squared. So three squared, so three squared is nine, multiplied by three is 27 X, which is what we have. And finally, we have minus B to the third. So it'd be three to the third and that is 27. So all the terms match up, which means that I can write the expression As X - To the 3rd power and notice how my two terms are now the same Of X -3. So I can combine those and write this as X -3 to the 4th power. So I can bring back the greater than or equal to zero. And from here, You can see that I have an expression X -3 raised to an even power. So because it's raised to an even power, I know that this expression is always going to be true. Meaning that say for example, I had negative one. So I have negative one to the 4th power because I have an even amount of negatives, this will always become positive one. So that sort of rule applies to our inequality, which means that X could be all real numbers. And if I write that in interval notation, X can span from negative infinity to positive infinity. So this becomes my final answer for this practice problem. And if we look at the answer choices, you can see that answer choice. A also spans from negative infinity to positive infinity. So hopefully this video helped and see you next time.

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. x⁴ + 6x² + 1 ≥ 4x³ + 4x

Key Concepts

Rearranging Polynomial Inequalities

Finding Critical Points by Solving Polynomial Equations

Testing Intervals and Using Interval Notation

Watch next

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. x4 + 6x2 + 1 ≥ 4x3 + 4x

Key Concepts

Rearranging Polynomial Inequalities

Finding Critical Points by Solving Polynomial Equations

Testing Intervals and Using Interval Notation

Watch next

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. x⁴ + 6x² + 1 ≥ 4x³ + 4x