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Finding roots in practice

Solving exercises about root factorization of polynomials
Hello. Welcome to “Finding roots in practice” step of week 3. Let us consider the first exercise. We have to find the root factorization of this polynomial. How can we do this? How can we solve this exercise? The idea is the following. We want to find the roots of this polynomial. In general, it’s not so easy. But you have seen in the Francis video a good algorithm to find rational roots of a polynomial, that is, roots of the form p/q, with p and q integers. Then the rational roots of this form of this polynomial
had to satisfy some conditions, precisely: you have that p has to divide the constant term of our polynomial. Minus or plus 4 is the same. And q has to divide the leading coefficient of our polynomial. Therefore, the possible candidates are the following. We have plus and minus 1/1, because 1 divides 4 and 1 divides 3. Then we have plus and minus 1/3, in that 1 divides 4 and 3 divides 3. Then we have plus and minus 2/1, plus and minus 2/3, plus and minus 4/1, and plus and minus 4/3. These are the possible candidates. Let us check for each one if they are or not roots of this polynomial. And let us start with 1.
If we substitute to x 1, what do we get? 3 minus 13 plus 16 minus 4. And this is equal– 3 minus 13 is minus 10, plus 16 is 6 minus 4 is 2, which is not 0. Therefore, 1 is not a root of our polynomial. And now we have to check minus 1. But observe, if you substitute to x in this polynomial any negative number, all the terms are negative. And therefore, you cannot get 0. You get a negative number. Therefore, this polynomial has no negative roots. Then it’s not necessary to check the negative candidates. Then let us now consider 1/3. What do we get?
We get 3 times 1 over 3 cubed minus 13 1 over 3 squared plus 16 times 1/3 minus 4. And this is equal to 1/9 minus 13/9– and it’s a good idea to take 9 as the common denominator now– plus 16 times 3 over 9 and minus 4 times 9 over 9. And what we get is– OK, 9 as common denominator– and then we have 1 minus 13, which is minus 12. Then 16 times 3 is equal 48.
And then we have minus 4 times 9, minus 36. And you’ll see the numerator is equal to 0, therefore, we get 0. Therefore, 1/3 is a root of our polynomial. And now let us consider the other positive candidate– 2.
If we substitute 2 to x, we get 3 times 2 cubed minus 13 times 2 squared plus 16 times 2 minus 4, which is equal to 3 times 8, which is 24, minus 13 times 4, which is equal to minus 52, plus 16 times 2, which is plus 32, minus 4. And now you see 24 minus 4 is 20, minus 52 plus 32 is minus 20. And therefore, we get 0. Then also 2 is a root of our polynomial. Observe, we have to check also the other candidates. But observe also now, what do we know? We know that our polynomial, which has degree 3, can be divided by x minus 1/3 times x minus 2.
This is because these two numbers are roots of this polynomial. And then a good idea to factorize this polynomial is to compute this product, to make the division of this polynomial by this one, and we get the last factor of degree 1. Observe that if this polynomial is divisible by this product, then it is divisible also by any scalar multiple of this polynomial. Therefore, it is divisible also by 3 times this one. I have multiplied by 3 just to eliminate this denominator and to have a simpler computation to do. Then let us compute this product, and we get 3x times x, which is 3x square.
Then we have 3 times minus 1/3 x, which is minus x minus 6x, therefore is minus 7x.
And finally, we have the product of these two, which is plus 2.
Then our polynomial is divisible by– our starting polynomial is divisible by this one. Let us compute the division. Then remember, we have to do the following. We’ll write here 3 x cubed minus 13 x squared plus 16 times x minus 4, and we have to divide by 3 x squared minus 7 times x plus 2. And we get x, 3 x squared times x is 3 x cubed minus 7 x squared plus 2 times x. We consider the subtraction, and we get minus 13 minus minus 7 is like minus 13 plus 7, which is minus 6 x squared, plus 14 times x minus 4.
Then we have to multiply 3 x squared with minus 2 to get minus 6 x squared. And then we write minus 2, and we consider the multiplication. And we get minus 6 x squared plus 14x minus 4. Consider the subtraction, and you get 0. Clearly, we get 0 as a remainder, because we knew in advance that our polynomial is divisible by this one. Then we can conclude that 3 x cubed minus 13 x squared plus 16x minus 4 is equal to 3 times x minus 1/3 times x minus 2 times, again, another copy of x minus 2. Thank you for your attention.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video.

In any case, you will find below a PDF file with the solutions.

Exercise 1.

Find the root factorization of the following polynomial:


Exercise 2. [Solved only in the PDF file]

Find the root factorization of the following polynomial: [3x^5+5x^4+4x^3+4x^2+x-1.]

Exercise 3. [Solved only in the PDF file]

Find the root factorizations of the following polynomials:

i) (x^4+2x^3-9x^2-2x+8 )

ii) (x^4-7x^3+17x^2-17x+6)

iii) (2x^4+x^3+3x^2+3x-9)

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Precalculus: the Mathematics of Numbers, Functions and Equations

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