For the given polynomial: a(x)=2x^5+5x^4+11x^3-49x^2-31x+26

Question

1. find the maximum possible numbers of the positive and the negative roots
2. find all the integer roots
3. find the non- integer real roots
4. find the complex roots

Answer 1

This is a huge task - let's take it step by step.

a. To find the maximum possible number of positive and negative roots, we need to use Descartes' Rule of Signs, which says that the number of sign changes in the coefficients of a(x) tells us the maximum # of positive roots, the number of sign changes in the coefficients in a(-x) tells us the maximum # of negative roots.

a(x) = 2x⁵ + 5x⁴ + 11x³ - 49x² - 31x + 26 changes signs twice (11 to -49 and -31 to 26), so the maximum number of positive roots is 2.

a(-x) = 2(-x)⁵ + 5(-x)⁴ + 11(-x)³ - 49(-x)² - 31(-x) + 26 = -2x⁵ + 5x⁴ - 11x³ - 49x² + 31x + 26 changes signs thrice, so the maximum number of negative roots is 3.

b. To find integer roots, we need to use synthetic division combined with the Rational Root Theorem. The theorem tells us that the rational roots will be a factor of the last term (26) divided by a factor of the first term (-2).

The factors of 26 are ±1, ±2, ±13, ±26, and the factors of 2 are ±1 and ±2. So our rational roots could be any of the first group divided by any of the second group (for example, 26, -13, -2, 1, etc. - so many possibilities).

Now we can apply synthetic division to test these combos. If we write out the coefficients of the polynomial:

2 5 11 -49 -31 26 we use a "sort-of long division" to test. If the end result is 0, we've found a root.

Let's try -1 first. It's quite difficult to show how this works here, but I'll do my best.

-1 | 2 5 11 -49 -31 26

Multiply the test (-1) by the first coefficient, then add it to the next one, then multiply again, and so on. Here are the steps one-by-one:

1. Start with the first coefficient: 2, then multiply (-1)(2) = -2
2. -2 + 5 = 3, (3)(-1) = -3
3. -3 + 11 = 8, (8)(-1) = -8
4. -8 - 49 = -57, (-57)(-1) = 57
5. 57 - 31 = 26, (26)(-1) = -26
6. -26 + 26 = 0

Since we ended up with 0, we know -1 is an integer root. Now we continue synthetically dividing, except we use the new coefficients I italicized above: 2, 3, 8, -57, 26. Once we synthetically divided by (x+1), we get a new polynomial:

2x⁴ + 3x³ + 8x² - 57x + 26

We can still use the possibilities that we came up with above. Let's try 2 as a possible root. Same thing, here are the steps:

2 | 2 3 8 -57 26

1. Start with 2 again, 2(2) = 4
2. 4 + 3 = 7, (7)(2) = 14
3. 14 + 8 = 22, (22)(2) = 44
4. 44 - 57 = -13, (-13)(2) = -26
5. -26 + 26 = 0

We got 0! So we now know 2 is a root. And we have another new polynomial with coefficients 2, 7, 22, -13.

c. 2x³ + 7x² + 22x - 13. Let's try another possible root: 1/2. By now you should be getting the hang of it:

1/2 | 2 7 22 -13

1. Start with 2 again, (2)(1/2) = 1
2. 1 + 7 = 8, (8)(1/2) = 4
3. 4 + 22 = 26, (26)(1/2) = 13
4. 13 - 13 = 0

Now we've got three real roots. We got -1, 2, and now 1/2. And now we've simplified the really hard-to-deal-with polynomial from the beginning to a parabola! We use the coefficients again to get:

d. 2x² + 8x + 26. Here, use your method of choice (quadratic formula is probably the easiest) to get the roots:

x = (-8±√¯64¯-¯4(2)(26)¯¯¯) / 4

= (-8 ± √¯-144¯ ) / 4 = -2 ± 3i.

And now (finally!), we've solved for all 5 roots:

-1, 2, 1/2, -2+3i, and -2-3i.

Answer 2

a(x) = 2x⁵+5x⁴+11x³-49x²-31x+26 --> changes sign twice

a(-x) = -2x⁵+5x⁴-11x³-49x²+31x+26 -> changes sign 3 times

1) 2 maximum positive roots and 3 maximum negative roots

2) Using rational root theorem

q = 2

p = 26

factors of p = ±(1,2,13)

factors of q = ±(1,2)

±(p/q) = ±(1,1/2,2,13,13/2)

Find a(±p/q) for which a(x) = 0

a(1/2) =0 yes

a(-1/2) ≠0 no

a(1) = -36 no

a(-1) = 0 yes

a(2) = 0 yes

a(-2) = -211 no

a(13) ≠ 0 no

a(-13) ≠ 0 no

Use synthetic division

2 5 11 -49 -31 26 |-1

-2 -3 -8 57 -26

-----------------------

2 3 8 -57 26 0

2 3 8 -57 26 |2

4 14 44 -26

-----------------

2 7 22 -13 0

2 7 22 -13 |1/2

1 4 13

--------------

2 8 26 0

Which leaves us with

2x²+8x +26 =0

Use quadratic equation

x = (-8 ± √(64 -208))/4 = -2 ± 6i

Integer roots are -1 and 2

3) Non-integer root is 1/2

4) Complex roots are -2+6i and -2-6i