Factor completely with respect to the integers.
1. 9x^2 - 4
2. x^3 + 64
3. 200x^2 - 50
4. 8x^3 - 64
5. x^3 + x^2 + x + 1
6. x^3 - 2x^2 + 4x - 8
7. 2x^3 + 4x^2 + 4x + 8
8. 2x^3 + 3x^2 -32x - 48
9. 7x^3 + 14x^2 + 7x
10. 6x^3 - 18x^2 - 2x +6
11. 3x^4 - 300x^2
12. x^4 + 12x^3 + 4x^2 + 48x
I need help solving these.
I need help on how to solve.
You must find the roots, which can be rational numbers, first. In simple cases this is not so difficult. In general you should use D'Alembert's theorem:
If the equation:
a_{n} x^n + a_{n-1} x^(n-1) + ...a_{1}x + a_{0} = 0
with all the a_{j} integers has a rational root x = p/q, p and q have no common factors (relatively prime), then p must divide a_{0} and q must divide
a_{n}. I give the proof below.
Problem 7:
Factor:
2x^3 + 4x^2 + 4x + 8
First take out a factor 2:
2(x^3 + 2x^2 + 2x + 4)
D'Alembert gives you the candidates for the rational roots of the polynomial inside the brackets. p must divide 4 so the possibilities are +/-1, +/-2
and +/-4. q must divide 1 so it can only be +/-1. This means that the rational roots can only be:
+/-1, +/-2 and +/-4
You find that -2 is a root. You then divide the polynomial by (x + 2) using e.g. long division to find a quadratic factor.
If none of the candidates you find using D'Alembert works you can conclude that the polynomial cannot be factored w.r.t. the integers provided the degree of the polynomial ios 3 or less. If a polynomial P_{N} of degree N can be factored then you have:
P_{N}= Q_{N1}Q_{N2}
where the Q_{N1} and Q_{N2} are polynomials of degree N1 and N2, respectively such that N1 + N2 =N.
For N =2 or 3 you see that you can only have a nontrivial factorization if one of the polynomials Q is of degree one.
If the degree is 4 then you could have two quadratic factors which don't contain linear factors. So, then you cannot conclude that such a polynomial cannot be factored if none of the rational candidates are roots.
In your case the fourth degree polynomial in (11) has a factor x^2 and in (12) it contains a factor x, so you can take these factors out and continue with the remaining quadratic and third degree polynomial.
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Proof:
Suppose you have an equation of the form
a_{n} x^n + a_{n-1} x^(n-1) + ...a_{1}x + a_{0} = 0
where all the a_{j} are integers. Suppose that a solution is x = p/q where p and q are integers that are relatively prime. Inserting x = p/q and multiplying by q^n gives:
a_{n} p^n + a_{n-1} p^(n-1)q + ...a_{1}pq^(n-1) + a_{0}q^n = 0
Let's write this as:
a_{n} p^n + a_{n-1} p^(n-1)q + ...a_{1}pq^(n-1) =- a_{0}q^n
Note that the left hand side is divisible by p:
[a_{n} p^(n-1) + a_{n-1} p^(n-2)q + ...a_{1}q^(n-1)]p = =- a_{0}q^n
The square brackets on the left hand side is an integer.
This then means that the right hand side must also be divisible by p. But q has no factors in common with p, so a_{0} must be divisible by p!
You can also easily see that a_{n} must be divisible by q by rewriting the equation as:
a_{n-1} p^(n-1)q + ...a_{1}pq^(n-1) + a_{0}q^n = -a_{n} p^n
You now observe that the left hand side is divisible by q, therefore the right hand side must also be divisible by q. Because p has no factors in common with q you can conclude that a_{n} must be divisible by q.
I have a 4th degree polynomial in which there are no rational zeros. I need to break it down into two quadratics but am unable to use the 'grouping method' since there is 5 terms. How do you get this to two quadratics?
3x^4-8x^3-5x^2+16x-5
Thanks
To factor the given 4th-degree polynomial into two quadratics, you can use the method called synthetic division or the rational root theorem.
1. Start by finding the possible rational roots using the rational root theorem. According to the theorem, the possible rational roots are divisors of the constant term (-5) divided by divisors of the leading coefficient (3). In this case, the possible rational roots are ±1/3, ±5/3, ±1, ±5.
2. Use synthetic division with each possible root to find any actual roots. Synthetic division is a method used to divide a polynomial by a binomial of the form (x - r), where r is a root of the polynomial. By performing synthetic division, you can determine if a possible root is indeed a root of the polynomial.
3. Once you find a root (let's say r), factor the polynomial by dividing it by (x - r). This will leave you with a quadratic factor.
4. Continue with the remaining factors obtained from synthetic division until you have factored the polynomial completely.
Let's apply this method to the given polynomial: 3x^4 - 8x^3 - 5x^2 + 16x - 5.
1. The possible rational roots are ±1/3, ±5/3, ±1, ±5.
2. Start with the possible root 1/3.
Using synthetic division:
| 3 -8 -5 16 -5
|_____ 1/3 -7/3 -12/3 4/3
|
The remainder is not 0, so 1/3 is not a root.
3. Move on to the next possible root, -1/3.
Using synthetic division:
| 3 -8 -5 16 -5
|_____ -1/3 3/3 2/3 -6/3
|________
3 -3 -2 8 -11
The remainder is not 0, so -1/3 is not a root.
4. Continue this process with the remaining possible roots until finding a root or exhausting all possibilities.
Since none of the possible roots are found to be actual roots, it means that this 4th-degree polynomial does not have any rational roots. In such cases, finding two quadratics is not possible. You may need to use numerical methods or factor using other techniques.