Which Cubics are Irreducible?

Find when the polynomial is reducible. If we write $f_{k}(x)$ as the product of two non-constant polynomials, we know one will have degree 1 and the other will have degree 2. $f_{k}(x)=x^{3}+x+k=(x+a)(x^{2}+bx+c)=x^{3}+(a+b)x^{2}+(c+ab)x+ac$ . Now we can match up the coefficients to get $a+b=0$ and $c+ab=1$ . Plugging in $a=-b$ , we get $c-a^{2}=1, c=1+a^{2}$ . We also have $k=ac$ , so $k=a+a^{3}$ . Remember that $a, b$ , and $c$ must be integers. $k$ is bounded by 1 and 1000, so we can see which values of $a$ produce valid values of $k$ . If $a \geq 10$ , then $k \geq 1010$ , so that doesn't work. All values of $a$ from 1 to 9 (inclusive) produce good values for $k$ , so we have 9 values of $k$ that allow for a reducible polynomial. Since there were 1000 total options, we have $1000-9=991$ which give an irreducible polynomial.

We'll count the number of REDUCIBLE polynomials, and subtract from 1000 to determine the number of irreducible polynomials.

$f_k(x)=x^3+x+k$ is a third-degree polynomial.

Since we are trying to break it down into the product of two polynomials of lesser degree, the sum of the degrees of the two polynomials must be 3 (since when they are multiplied together, the result is a degree 3 polynomial). Additionally, for the original polynomial to be reducible, each of the two new polynomials must have degree of at least one. The only possible pair of degrees that meet these requirements are 1 and 2.

Thus, we can write $f_k(x)$ in "reduced" form as the product of the following two generic polynomials of degree 1 and 2: $(x+a)(x^2-bx+k/a)$ .

Expanding this product, we have $x^3+ax^2-bx^2-abx+\frac{k}{a}x+k=x^3+(a-b)x^2+(\frac{k}{a}-ab)x+k$ .

Since there is no $x^2$ term in the original polynomial, $a-b$ must equal 0, so $a=b$ .

Replacing each $b$ with $a$ , our polynomial becomes $x^3+(\frac{k}{a}-a^2)x+k$ . Since the coefficient on $x$ is 1 in the original polynomial, we have $\frac{k}{a}-a^2=1$ .

Solving for $k$ , we have $k=a^3+a$ .

Now, we must identify values of $a$ that generate $k$ values such that $1\leq k\leq1000$ .

$a$ must be an integer in order for $k$ to be an integer. If $a\leq0$ , then $k<1$ , so we exclude those $a$ values. If $a\geq10$ , then $k>1000$ , so we exclude those values as well. This leaves $1\leq a\leq9$ , which generate a total of 9 values for $k$ .

Letting $a$ increase from 1 through 9, the $k$ values that are generated are 2,10, 30, 68, 130, 222, 350, 520, 738.

Since there are 9 values $k$ that cause of $f_k(x)$ to be reducible, there are $1000-9=981$ values of $k$ such that $f_k(x)$ is irreducible.

Polynomials of this form always have one real zero, because an $x^2$ term or negative x term must be present for there to be two or three real zeros. The polynomial would have to pass through 0 and then move back down to 0. A negative slope is required for this, but in the derivative, $3x^2$ is always positive, the second term is 1, and the third term disappears. For the polynomial to be reducible, the real zero must be an integer. $x^3+x=-k$ , x is an integer, and k is between 1 and 1000. For $x \geq 0$ , $k \leq 0$ , and for $x \leq -10$ , $k>1000$ . The only valid x values are those from 1 to 9. Each of these corresponds to a single integer k. The remaining 991 k values create irreducible polynomials.

First we find all $1\leq k\leq 1000$ so that $x^3+x+k$ is reducible. Let $x^3+x+k=(x^2+ax+b)(x+c)=x^3+(a+c)x^2+(ac+b)x+bc$ where $a,b,c$ are integers. Since the coefficient of $x^2$ is $0$ , $a=-c$ . Since the coefficient of $x$ is $1$ , $ac+b=-c^2+b=1$ implies that $b=c^2+1$ . We have $k=bc=c^3+c$ and $1\leq k\leq 1000$ implies that $1\leq c\leq 9$ . Since all these $k$ are distinct, so $k$ has $9$ values for $x^3+x+k$ to be reducible. So the other $1000-9=991$ other values of $k$ make $x^3+x+k$ irreducible.

We find all the values of k when x³+x+k is reducible and subtract from 1000.

If x³+x+k is reducible over the polynomials with integer coefficients, then it is factorable as the product of a linear binomial and a quadratic trinomial both with leading coefficients 1. That is, integers A,B,C exist such that:

x³+x+k = (x+A)(x²+Bx+C)

Multiplying the right side out we get

     (x+A)(x²+Bx+C) = x³+(A+B)x²+(AB+C)x+AC

So we have the identity:

x³+(A+B)x²+(AB+C)x+AC = x³+x+k

So we can equate coefficients:

(1) A+B= 0 (2) AB+C = 1 (3) AC = k

So from (1), B = -A, and substituting for B in (2),

A(-A)+C = 1 -A²+C = 1 (4) C = A²+1

So the factorization x³+x+k = (x+A)(x²+Bx+C) becomes

x³+x+k = (x+A)(x²-Ax+A²+1)

Substituting for C from (4), in (3),

A(A²+1) = k A³+A = k

Since 1 ≤ k ≤ 1000 1 ≤ A³+A ≤ 1000

A³+A is a strictly increasing function, therefore:

The minimum value of A is 1, when k = A³+A = 1³+1 = 1+1 = 2, and The maximum value of A is 9, when k = A³+A = 9³+9 = 729+9 = 738. (For when A = 10, k = A³+A = 10³+10 = 1000+10 = 1010, which is over 1000.)

So there are 9 values of A, 1 through 9, and therefore 9
factorizations which are:

For k = 1, (x³+x+2) =(x+1)(x²-x+2) For k = 2, (x³+x+10) =(x+2)(x²-2x+5) For k = 3, (x³+x+30) =(x+3)(x²-3x+10) For k = 4, (x³+x+68) =(x+4)(x²-4x+17) For k = 5, (x³+x+130) =(x+5)(x²-5x+26) For k = 6, (x³+x+222) =(x+6)(x²-6x+37) For k = 7, (x³+x+350) =(x+7)(x²-7x+50) For k = 8, (x³+x+520) =(x+8)(x²-8x+65) For k = 9, (x³+x+738) =(x+9)(x²-9x+82)

Since for only 9 positive integers 1 ≤ k ≤ 1000, the polynomial f k(x)=x³+x+k is reducible, then for the other 1000-9 or 991 positive integers 1 ≤ k ≤ 1000, the polynomial f k(x)=x³+x+k is irreducible.

Answer is 991

Assume the polynomial is reducible, in meaning, can be divided by x + a, where a is an integer.

x^3 + x + k = ( x + a ) ( x^2 - ax + 1 - a^2 )

By performing long division, this is the result. At the final stage of division, it can be shown that k = a( 1 - a^2 ). And for k less than 1000, only a= 1,2,3,...,9 satisfy the equation. So there is 9 values of k that makes equation divisible (reducible) into product of two polynomials with lower degree. And hence

No of k that makes equation irreducible = 1000 - 9 -991

If the given equation is reducible, by Ferrari's method, assume the following: x^3+x+k = (x+p)(x^2+qx+r) where p,q & r are integers. =>x^3+x+k = x^3+(p+q)x^2+(pq+r)x+pr Comparing coefficients we get p+q = 0 , pq+r = 1 , pr = k i.e. k = p(1+p^2) , r = 1+p^2 Since 1<=k<=1000 , we get the following values of p from k = p(1+p^2): p = 1,2,3,4,5,6,7,8,9 and correspondingly k=2,10,30,68,130,222,350,520,738 Thus for the above 9 values of k the given equation is reducible. So the equation is irreducible for 1000-9 i.e. 991 values of k.

let x^3+x+k=(x^2+a x+b) (x+c) R-H-S x^3+(a+c) x^2+(b+a c) x+b c; now comparing the two equations we get a+c=0=>a=-c; b+c a=1; "a=-c" b-a a=1; b c=k; b a=-k; but b=1+a a; and (1+(-a) (-a))*(-a)=k hence we can just check that for -10<a<0 we have net 9 values in which it can be formatted hence net are 1000-9==991

Assuming that the given polynomial has 1 integer root r we can deduce the following:

$x^3+x+k=(x-r)(x^2+ax+b)=x^3+(a-r)x^2+(b-ar)x-br\iff$

$\iff\ a=r$ , $b-ar=1$ , $-br=k$ $\iff\ a=r$ , $b=1+r^2$ , $br=-k$

Now we know that $f_k(x)$ has an integer root r, iff a,b are integers.Now that there exists a 1-1 relationship between $f_k(x)$ with integer roots and reducible $f_k(x)$ ,all we have to do in order to find all reducible $f_k(x)$ is let r assume integer values and then check if br<1000. Since br<0 and b>0 for all r, we may only let r<0. Checking up to r=-10 we find that k=1010, so we reject every polynomial for which r<=-10 since k=-br is a monotonously increasing function of r. Thus the only reducible polynomials with $1\leq{k}\leq{1000}$ are 9, those with $r={-1,-2,...,-9}\implies{k={2,10,...,738}}$ . All the other 1000-9=991 values of k produce irreducible polynomials, which is the final answer.

This problem is best solved by first counting the number of positive integers k that makes x^3 + x + k reducible (can be expressed as product of two other polynomials with integer coefficients). Assuming it is divisible by x + N, where N is an integer.

x^3 + x + k = ( x + N )( x^2 - Nx + [1 + N^2] )

We know that k = N ( 1 + N^2 ), where k is positive integer in range of 1 to 1000 that makes polynomial reducible.

So right now, we have N = 1,2,3,...,9 that results in value of integer k less than 1000. Hence there is 9 values of integer k which make the polynomial reducible. From 1 to 1000 (1 and 1000 inclusive), there are 9 values of k that makes polynomial reducible. Hence, the number of integer values k which make polynomial IRREDUCIBLE is:

Number of values k = 1000 - 9 = 991

And Hence the solution.

Let $a$ and $b$ be integers such that $x^3+x+k=(x+a)(x^2-ax+b)$ . Then by Vieta's formulas, we get $-a^2+b=1, ab=k$ . It follows that $k=f(a)=a(a^2+1)=a^3+a$ . Note that $k>a^3$ and $1\leq k\leq 1000$ , we need to consider the values of $k$ for $a=1,2,3,\cdots 9$ . These give $9$ values of $k$ , so there are $9$ reducible polynomials for $1\leq k\leq 1000$ , and $1000-9=991$ irreducible polynomials.

f(x)=x^3+x+k note in f(x) k>0 thus for f(x) to have a factor x^3+x<0,which is true only when x<0 since k is a integer,x must also be a integer and also k<=1000 on imposing these conditions, we can get 9 value of k for which f(x) is reducible.thus for 991 value of k ,f(x) is irreducible

1 < or = k = (-x^2)(x+1) < or = 1000. x = -1,-2,-3...,-9 work. done.

since this is a cubic, one of the factors has to be (x+m) where m = integer..

x=-1,……-9

A polynomial $f_k$ is reducible if and only if it is a product of two nontrivial polynomials with integer coefficients. Because the degree of $f_k$ is three, one of these polynomials has to be linear. Because the polynomial $f_k$ is monic (i.e., the highest coefficient is $1$ ), this linear polynomial must have the form $(x-a)$ for some integer $a$ . Thus, the polynomial $f_k$ is reducible if and only if it has an integer root. Note that $f_k(a)=0$ is equivalent to $a^3+a=-k.$ The function $a^3+a$ is increasing, and one can easily check that $a^3+a$ belongs to $[-1000,-1]$ if and only if $a$ is from $-9$ to $-1$ . So the number of reducible polynomials with $1\leq k \leq 1000$ is $9$ and the answer is $1000-9=991$ .

We have f(x) = x^3 + x + k From here we will find the number of k that make the given equation a reducible one. It follows that: x^3 + x + k=0 -> x^3 + x = -k -> x( x^2 + 1 ) = -k Then we will find those integer values of x that limit the value of k from 1 to 1000. we will have the values of x= { -1, -2, -3, -4, -5,-6, -7.-8,-9} We can now conclude that there are 9 reducible forms of the given equation. And 991 irreducible forms.