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Hello, and welcome back to a step in practice. We are dealing today with polynomial inequalities. And in this video, we’re going to solve just the first exercise. You’ll find the second one in the PDF. We want the polynomial to be strictly negative. But they told us that minus 1 is a root. Actually, a double root. This means that we can divide this polynomial by x plus 1 to the square. So we are going to perform the division of the polynomial with x plus 1 to the square. x plus 1 to the square is x square plus 2x plus 1.

58.2

Now in order to get x to the fourth, I have to multiply x square by x square. And if I multiply x square with x square plus 2x plus 1, I get x to the fourth plus 2x to the cube plus x square. And so the difference between the two polynomials will be 0x to the fourth, then minus 5x to the cube minus 4x square plus 7x plus 6. In order to get minus 5x to the cube, I have to multiply x square by minus 5x. But if I multiply minus 5x with x square plus 2x plus 1, what we get is minus 5x to the cube minus 10x square minus 5x.

110.1

And so the difference between these two polynomials will be, well, 0x to the cube, and then 6x to the square plus 12x plus 6, which is actually 6 times x square plus 2x plus 1. So in order to get this polynomial, it is enough to multiply by 6 x square plus 2x plus 1. And then we get remainder 0. So the initial polynomial is the product of x square plus 2x plus 1, with x square minus 5x plus 6. Now, let us find the factors of x square minus 5x plus 6. Well, the discriminant of this polynomial is 5 to the square minus 24, which is 1.

165.9

And therefore, it has two roots, 5 plus or minus 1 over 2, which gives us 3 and 2. Therefore, the initial polynomial can be decomposed into the following factors. So x to the fourth minus 3x cube minus 3x square plus 7x plus 6 is equal to x square plus 2x plus 1, which is x plus 1 to the square times, well, x minus 3, x minus 2. Or equivalently, here we can write x square minus 5x plus 6. And we recall that the roots of this polynomial are 2 and 3. Now let us study the sign of this product. And we use the rule sign.

222.5

So we write the factors here, x, and here we write x plus 1 to the square. And below, we write x square minus 5x plus 6. And here, we write the product, which is our initial polynomial. Now, the values that count here are minus 1, 2, and 3. So let us write here minus 1, 2, and 3. Now, x plus 1 to the square vanishes in minus 1, and is positive x elsewhere, whereas x square minus 5x plus 6 vanishes in 2 and 3. Recall it is a polynomial of second degree, and thus it is– since the coefficient of x square is 1, it will be negative inside– in the interval between the two roots, and positive elsewhere.

282.7

So the product of the two factors is positive here, positive here, negative here, and positive here. Zero, zero, and zero. Now, we are looking for the polynomial to be strictly negative. So it will be strictly negative just in this interval, except its boundaries. So the solution will be the open interval 2, 3. And this ends exercise one. The next exercise will be solved the PDF. So see you in the next step. Bye.