How do you find the rational roots of each polynomial equation?

Question

I'm seriously desperate for help. I've done every single homework problem but am stuck on these three problems on rational roots. Please help me :(

1) 4x^3 + 3x^2 + x + 2 = 0

2) 3x^3 -2x^2 -12x + 8 = 0

3) 4x ^3 -13x^2 + 11x - 2 = 0

Answer 1

1) Put 1 in for x to see if the equation equals zero. If it doesn't then try -1. What I usually do is 1,-1,2,-2,... it's not very often I have to go beyond that. Anyway when you put -1 in for x the equation equals zero so that means (using remainder theorem) that -1 is a solution to the equation. Then use that -1and synthetic division to get 4x^2-x+2=0, you can't factor this and get real solutions. The discriminant, b^2-4ac, tells you that so the only rational root is -1.

2) Factor by grouping.
x^2(3x-2)-4(3x-2)=0
(3x-2)(x^2-4)=0
(3x-2)(x+2)(x-2)=0
Rational roots: x=2/3, x=-2, x=2

3) Use the same technique as you did in problem
1. You'll find that 1 is a solution so use it and synthetic division. You get 4x^2-9x+2=0 which factors into (4x-1)(x-2)=0
So the rational roots are: x=1, x=1/4, x=2

Answer 2

Rational root theorem
From Wikipedia, the free encyclopedia

In algebra, the rational root theorem states a constraint on solutions (also called "roots") to the polynomial equation

an xn + an−1 xn −1 + ... + a1 x + a0 = 0
with integer coefficients. Let an be nonzero. Then each rational solution x can be written in the form x = p/q for p and q satisfying two properties:

p is an integer factor of the constant term a0, and
q is an integer factor of the leading coefficient an.
Thus, a list of possible rational roots of the equation can be derived using the formula x = ± p/q.

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Problem 1:

So, given this hint from wikipedia, try some roots:

Note none of the positive roots are possible (all would give a
sum greater than zero)

Try -1: -4 + 3 - 1 + 2 = 0 is true so -1 is a root.

Now factor the polynomial above :
( x + 1 ) ( 4 x^2 - x + 2)

Are there any rational factors for the second term?
Use the discriminent for the quadratic equation:

+ - sqr(b^2 - 4ac) = + - sqr (1 - 4 * 4 * 2)

With the stuff in the sqr sign negative, there are no more real roots.

The root is x = -1
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Answer 3

The Rational Root Theorem says that a polynomial with integer coefficient has a rational root p/q (in lowest words) provided that p is a divisor of the consistent coefficient and q is a divisor of the maximum popular coefficient. on your polynomial, meaning that p might desire to be a divisor of 9 and q might desire to be a divisor of four. the two beneficial and unfavourable divisors count kind, so as meaning: p might desire to be in (a million, -a million, 2, -2, 4, -4) q might desire to be in (a million, 3, 9) that provides 18 opportunities to attempt: for each of 6 distinctive p values, there are 3 distinctive q values to attempt. of course, because of the fact that's a cubic polynomial, at maximum 3 distinctive roots exist. additionally, the Rational Root Theorem would not assure any rational roots in any respect. It purely tells you which of them rational numbers ought to probable be roots. to your polynomial, those are a million, -a million, a million/3, -a million/3, a million/9, -a million/9, 2, -2, 2/3, -2/3, 2/9, -2/9, 4, -4, 4/3, -4/3, 4/9, -4/9

Answer 4

If p/q is a primitive fraction [that is p and q have no common divisor] and p/q is a root of a polynomial of the form

A0 x^n + A1 x^{n-1} + ... + An,

then q is a divisor of A0 (lets write this as q|A0) and p is a divisor of An (write p|An).

You can easily figure this out by substituting x = p/q into the polynomial and doing some simple observations [I leave details out, if you need them, ask].

About your problems:

1) q|4 and p|2; so you have these options for q = {-4,-2,-1,1,2,4} and these for p = {-2,-1,1,2}.

Now try all 24 (6 times 4) combinations of p/q to check whether you got a root or not. All solutions you get are all that exist.

You can do the same for problems 2 and 3; I hope the concept is clear.

Hope this helps!

Answer 5

Well smd, cubics are hard to factor by inspection, so you may have to go to trial and error. The equations can guide you, though.

1. Eqtn indicates a negative root at x=-1
2. Eqtn indicates a root between 0 and 1 and between -2 and -3.
3. Eqtn indicates a root at x=1 and a negative root.