# Definition For a set F, a polynomial over F with variable x is of the form

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1 *7. Polynomials Definition For a set F, a polynomial over F with variable x is of the form a n x n + a n 1 x n 1 + a n 2 x n a 1 x + a 0, where a n, a n 1,..., a 1, a 0 F. The a i, 0 i n are the coefficients of the polynomial. If x n is the largest power of x appearing in the polynomial then n is the degree of the polynomial, a n x n is the leading term and a n is the leading coefficient. The collection of all polynomials with one variable x and with coefficients from F will be denoted by F [x]. (Note the square brackets.) Note that 0 F [x], being x 2 + 0x + 0, but it is not said to have a degree, though some books give it degree 1 or even. Part of the aim of this section is to show how similar are the properties of F [x] and Z. Examples 3x 2 + 5x 1 Z [x], 3 7 x x2 + x Q [x], x 2 π R [x]. We could also look at polynomials in Z m [x] or C [x]. If we can add and multiply numbers in the set F then we can add and multiply the polynomials in F [x]. Examples (i) In Z [x] the sum of 3x 2 + 5x 1 and 5x 3 3x 2 + 2x + 1 is ( 3x 2 + 5x 1 ) + ( 5x 3 3x 2 + 2x + 1 ) = 5x 3 + 7x. (ii) In Z [x] the difference of x and 9x + 7 is ( x ) ( 9x + 7 ) = x 2 9x 6. Notice how subtraction is easier than for integers. For example is complicated by the fact that we can subtract 7 from 1 and put 6 in the units place, we have to instead borrow 10 from the next column. 1

2 (iii) In Z [x] the product of x 2 + 2x + 3 and x is ( x 2 + 2x + 3 ) ( x ) = 3x x 3 + 8x +x 4 + 4x 2 = x 4 + 2x 3 + 7x 2 + 8x + 12 We can do all the above again in more interesting sets. (i) Addition in Z 3 [x]. The sum of 2x 3 + 2x 2 + x + 1 and x 3 + 2x is using 3 0 mod 3 and 4 1 mod 3. 2x 3 + 2x 2 + x ( x 3 + 2x ) = x 2 + x, is (ii) Subtraction in Z 5 [x]. The difference of 4x 3 +3x 2 +x+3 and 2x 3 +3x+4 ( 4x 3 + 3x 2 + x + 3 ) ( 2x 3 + 3x + 4 ) = 2x 3 + 3x 2 + 3x + 4 using 2 3 mod 5 and 1 4 mod 5. (iii) Multiplication in Z 2 [x]. The product of x 3 + x + 1 and x 2 + x + 1 is ( x 3 + x + 1 ) ( x 2 + x + 1 ) = x 5 + x 4 + x 3 using 2 0 mod 2. + x 3 + x 2 + x + x 2 + x + 1 = x 5 + x Note that in Z [x], Q [x], R [x], C [x] or Z p [x], with p prime, we have deg fg = deg f + deg g. This may not hold in Z m [x] with m composite. Example In Z 6 [x] multiply f (x) = 2x and g (x) = 3x to get ( 2x ) ( 3x ) = 6x 6 + 3x 4 + 2x = 3x 4 + 2x Hence, in this example, deg fg < deg f + deg g. 2

3 Subtle point. Recall that two functions f and g are equal on a set X if, and only if, f (x) = g (x) for all x X. Consider f (x) = x 4 + 2x and g (x) = (1 + x) 2 over Z 3. There are only three elements in Z 3 and for these, f (0) = 1 = g (0), f (1) = 1 = g (1), f (2) = 0 = g (2). So as functions over Z 3 these are equal though as polynomials they are different. Question If we can multiply polynomials can we factor them? Assumption α F, α 0, α 1 F. So, from now on F will be one of Q, R, C or Z p, with p prime because, to look at factorization, or dividing, it is best to restrict to sets F in which we can find inverses. (We can talk of dividing in Q, R or C but not in Z p. Instead we have to remember that to divide by something is to multiply by it s inverse.) (Aside: In Z 6 does [3] 6 have an inverse? i.e. does there exist an element with [3] 6 [x] 6 = [1] 6. If there did, we could multiply both sides by [2] 6 to get [6] 6 [x] 6 = [2] 6, i.e. [0] 6 = [2] 6, contradiction. So not every element in Z 6 has an inverse. The same argument works in general for Z m with m composite.) The first step to answering this question about factoring is to look at long division for polynomials. Example In Q [x] divide x 2 x 1 into x 4 + 2x 3 + 3x 2 + 4x + 5. Solution x 2 + 3x + 7 x 2 x 1 ) x 4 + 2x 3 + 3x 2 + 4x + 5 x 4 + x 3 + x 2 3x 3 + 4x 2 + 4x 3x 3 + 3x 2 + 3x 7x 2 + 7x + 5 7x 2 + 7x x + 12 Hence x 4 + 2x 3 + 3x 2 + 4x + 5 = (x 2 + 3x + 7) (x 2 x 1) + (14x + 12). 3

4 Always, always check by multiplying out. You should never end with an incorrect answer. Thus given f and g Q [x] we have found q, r Q [x] such that f = qg+r and deg r < deg g. This is very reminiscent of the division theorem for integers and we have exactly the same result for polynomials. Example In Z 3 [x] divide x 2 + x + 1 into x 4 + x x 2 + 2x + 1 x 2 + x + 1 ) x 4 + x x 4 + x 3 + x 2 2x x 3 + 2x 2 + 2x x 2 + x + 1 x 2 + x (Often using 2 = 1, 1 = 2 modulo 3.) Hence x 4 + x = (x 2 + 2x + 1) (x 2 + x + 1) and so x 2 + x + 1 divides x 4 + x Definition If f, g F [x] we say that g (x) divides f (x), and write g f, if there exists q (x) F [x] such that f (x) = g (x) q (x). We also say that f if a multiple of g. Note that if g f then either g 0 (we say it is identically zero) or deg g deg f. Careful If m, n Z, then m n and n m imply m = ±n. If we then demand that both m and n are positive we get m = n. But if we have polynomials f, g F [x] then f g implies deg f deg g. Similarly, g f implies deg g deg f. Hence deg f = deg g. But f g also means f = gu for some polynomial u. Then deg f = deg g means deg u = 0, and so u is a non-zero constant. Hence f g and g f imply only that f = cg for some constant c. Definition If f F [x] can be factored f (x) = g (x) h (x), with g, h F [x] and 1 deg g, deg h < deg f, then we say f is reducible. If we cannot factor f in this way we say it is irreducible. Compare this to the definition of prime numbers. It can be hard to check if a given polynomial is irreducible. And it can depend on the set F, see below. 4

5 Theorem Division Theorem for Polynomials. Let F be as above. Let f (x), g (x) F [x] with g (x) 0. Then there exist unique polynomials q (x), r (x) F [x] such that f (x) = q (x) g (x) + r (x) and either r (x) 0 or r is non-zero and has a lower degree than g. Proof Not given in course, but see appendix. An important deduction connects the roots of a polynomial with factorization. Corollary Let F be as above. Let p (x) F [x]. Then p (a) = 0 if, and only if, (x a) p (x). Proof ( ) Assume p (a) = 0. Apply the Theorem with f = p and g = x a to find q, r F [x] with p (x) = q (x) (x a) + r (x), and either r 0 or deg r < deg g. If r 0 we are finished. Otherwise deg r < deg g = 1 and so deg r = 0, i.e. r is a constant, c say. Thus p (x) = q (x) (x a) + c. Put x = a to see that c = 0. So again x a divides p (x). ( ) Assume (x a) p (x). But this means p (x) = q (x) (x a) for some q (x) F [x]. Putting x = a and we see that p (a) = 0. Since a polynomial in F [x] of degree n can be the product of at most n linear factors x a, it can have at most n roots in F. But will it have exactly n roots? Example x 2 2 Q [x] yet it has no roots in Q. Its roots are ± 2 R. Further x R [x] but it has no roots in R. Its roots are ±i C. Something different happens for C. Theorem Fundamental Theorem of Algebra. If f C [x] has deg f 1 then f has a root in C. Thus f factorizes completely into n linear factors. Proof Not given in this course. So a polynomial in C [x] of degree n has exactly n roots in C. This means that the only irreducible polynomials in C [x] are the linear polynomials, ax + b. The Corollary can help us factorize polynomials; to find linear factors it suffices to find roots. 5

6 Examples (i) Factorize p (x) = x 4 + x 3 2x 2 6x 4 over R. Solution Look at p (a) for small a. So p (0) = 4, p (1) = 8, p ( 1) = 0. Thus x + 1 divides p (x). x 3 2x 4 x + 1 ) x 4 + x 3 2x 2 6x 4 x 4 + x 3 2x 2 6x 4 2x 2 2x 4x 4 4x 4 0 x 4 + x 3 2x 2 6x 4 = (x + 1) ( x 3 2x 4 ). Writing q (x) = x 3 2x 4 we see that q (2) = 0 so x 2 divides q (x). x 2 + 2x + 2 x 2 ) x 3 2x 4 x 3 2x 2 2x 2 2x 4 2x 2 4x 2x 4 2x 4 0 x 3 2x 4 = (x 2) ( x 2 + 2x + 2 ). Does x 2 +2x+2 factorize over R? Note that x 2 +2x+2 = (x + 1) and so can never be zero for real x. Thus x 2 + 2x + 2 has no linear factors. Hence we finish with x 4 + x 3 2x 2 6x 4 = (x + 1) (x 2) ( x 2 + 2x + 2 ). (ii) Factorize p (x) = x 4 + x 3 2x 2 6x 4 over Z 5. The polynomial is the same as in (ii) so we still have But now, x 4 + x 3 2x 2 6x 4 = (x + 1) (x 2) ( x 2 + 2x + 2 ). x x 2 + 2x + 2 mod

7 Thus modulo 5, x 2 +2x+2 has roots at 1 and 2. So, in Z 5 [x], x 2 +2x+2 = (x 1) (x 2) = (x + 4) (x + 3). (Check this by multiplying out, modulo 5.) Hence, in Z 5 [x], x 4 + x 3 2x 2 6x 4 = (x + 1) (x 2) (x + 4) (x + 3) = (x + 1) (x + 3) 2 (x + 4). Aside So here we have seen that x 2 +2x+2 is irreducible in R [x] yet reducible in Z 5 [x] And of course it is reducible in C [x] : x 2 + 2x + 2 = (x + 1) = (x + 1) 2 i 2 = ((x + 1) + i) ((x + 1) i), difference of squares = (x + 1 i) (x i). (iii) Factorize p (x) = x 4 + x 3 + x 2 + 3x + 4 over Z 5. Solution We need only check x = 0, 1, 2, 3 2 or 4 1 mod 5. In turn, p (0) = 4, p (1) = 0, p (2) = 3 while p ( 2) = 0 and p ( 1) = 2, all calculation modulo 5. Thus x 1 and x + 2 divide p (x). So for some q (x) we have p (x) = (x 1) (x + 2) q (x) = (x 2 + x 2) q (x), i.e. x x 2 + x 2 ) x 4 + x 3 + x 2 + 3x + 4 x 4 + x 3 2x 2 3x 2 + 3x + 4 3x 2 + 3x x 4 + x 3 + x 2 + 3x + 4 = (x 1) (x + 2) ( x ). You should check that x has no zeros. If p (x) has repeated roots then x could have 1 or 2 as zeros, just as p (x) did. But x x mod Hence x 4 + x 3 + x 2 + 3x + 4 = (x 1) (x + 2) ( x ) = (x + 4) (x + 2) ( x ) in Z 5 [x]. 7

8 The division Theorem above is very similar to a result in integers. Other ideas can be generalized to polynomials such as Definition Let F be one of Q, R, C or Z p, with p prime. Let f, g F [x] be polynomials not both identically zero. Then a greatest common divisor is a polynomial d F [x] such that (i) d (x) divides both f (x) and g (x), (ii) if c (x) divides both f (x) and g (x) then c (x) divides d (x). Note that if the gcd exists it is not unique. For if d satisfies (i) and (ii) and λ F is non-zero then λd also satisfies the two properties. For two polynomials f and g write gcd (f, g) to represent any greatest common divisor. A particular gcd is often picked out by taking λ = c 1 where c is the leading coefficient of d. Then c 1 d has leading coefficient 1, and is called a monic polynomial. We need to justify that such a gcd exists. argument as used for the gcd of integers. We do not use the same Theorem Let F be as above. Let f, g F [x] be polynomials not both identically zero. Then the gcd (f, g) exists. Proof Not given in course but see the appendix. In any given problem we construct a gcd using repeated use of the Division Theorem. And if we formalize this method we get Theorem The Euclidean Algorithm for Polynomials. Let F be as above. Let f, g F [x] be polynomials. If g divides f then g is a gcd of f and g. Otherwise apply the Division Theorem to obtain a series of equations f = gq 1 + r 1, 0 < deg r 1 < deg g, g = r 1 q 2 + r 2, 0 < deg r 2 < deg r 1, r 1 = r 2 q 3 + r 3, 0 < deg r 3 < deg r 2,. r j 2 = r j 1 q j + r j, 0 < deg r j < deg r j 1, r j 1 = r j q j+1. Then r j, the last non-zero remainder, is a greatest common divisor of f and g. Further, by working back up this list we can find p (x), q (x) F [x] such that gcd (f (x), g (x)) = p (x) f (x) + q (x) g (x). Proof Left to student. 8

9 Examples (i) Over Q [x] find a greatest common divisor of f (x) = x 5 + 3x 4 + 5x 3 + 5x 2 + 4x + 2 and g (x) = x 4 + 2x 3 + 4x 2 + 4x + 4 and write the answer as a linear combination of f and g. Solution Applying the division algorithm we get x + 1 x 4 + 2x 3 + 4x 2 + 4x + 4 ) x 5 + 3x 4 + 5x 3 + 5x 2 + 4x + 2 x 5 2x 4 4x 3 4x 2 4x x 4 + x 3 + x x 4 2x 3 4x 2 4x 4 x 3 3x 2 4x 2 x + 1 x 3 3x 2 4x 2 ) x 4 + 2x 3 + 4x 2 + 4x + 4 x 4 3x 3 4x 2 2x x 3 + 2x + 4 x 3 + 3x 2 + 4x + 2 3x 2 + 6x x 1 3 3x 2 + 6x + 6 ) x 3 3x 2 4x 2 x 3 + 2x 2 + 2x x 2 2x 2 x 2 + 2x So we have found a zero remainder. The last non-zero remainder can be taken as the greatest common divisor i.e. 3x 2 + 6x + 6. We can work back and write the gcd as a linear multiple of f and g. So r 2 (x) = g (x) ( x + 1) ( x 3 3x 2 4x 2 ) = g (x) ( x + 1) (f (x) (x + 1) g (x)) = (1 + ( x + 1) (x + 1)) g (x) ( x + 1) f (x) = ( 2 x 2) g (x) + (x 1) f (x). Always, always check your answer by multiplying out: ( 2 x 2 ) ( x 4 + 2x 3 + 4x 2 + 4x + 4 ) + (x 1) ( x 5 + 3x 4 + 5x 3 + 5x 2 + 4x + 2 ) = 3x 2 + 6x + 6 9

10 (ii) Over Z 3 [x] find a greatest common divisor of x 3 + 2x 2 + 2x + 1 and x and express your answer as a linear combination of the original polynomials. Solution Firstly Next x = x ( x 3 + 2x 2 + 2x + 1 ) + x 3 + x 2 + 2x + 2 = (x + 1) ( x 3 + 2x 2 + 2x + 1 ) + 2x x 3 + 2x 2 + 2x + 1 = 2x ( 2x ) + 2x = (2x + 1) ( 2x ). So gcd (x 3 + 2x 2 + 2x + 1, x 4 + 2) = 2x And, quite simply, 2x = ( x ) (x + 1) ( x 3 + 2x 2 + 2x + 1 ). 10

11 Appendix 1) Theorem Division Theorem for Polynomials. (H.P p. 264) Let F be as above. Let f (x), g (x) F [x] with g (x) 0. Then there exist unique polynomials q (x), r (x) F [x] such that f (x) = q (x) g (x) + r (x) and either r (x) 0 or r is non-zero and has a lower degree than g. Proof (Existence) If f (x) 0 we can write 0 = 0g (x)+0, so we can assume that f (x) 0. If g f then f (x) = g (x) q (x) for some q (x) and the result follows with r (x) 0. In particular if deg g = 0, i.e. g (x) c a non-zero constant, we can take q (x) = c 1 f (x). Hence we can assume that g f and n = deg g 1. Consider the set S = {f qg : q F [x]}. Because g f this set does not contain 0 so every element in this set has a degree. This set is non-empty, since it contains f 0g. If we look at D, the set of degrees of the polynomials in S, we have a non-empty set of nonnegative integers so, by the well-ordering principle, it contains a minimum element. Choose a q F [x] such that f qg takes this minimum value and set r = f qg. Since g f we have r (x) 0 and so we have to show that deg r < deg g. Assume for the sake of contradiction that deg r deg g. Let m = deg r and write r (x) = a m x m + a m 1 x m a 1 x + a 0, along with g (x) = b n x n + b n 1 x n b 1 x + b 0. So m n, a m 0 and b n 0. Consider the polynomial r (x) a m b 1 n x m n g (x) = (f (x) q (x) g (x)) a m b 1 n x m n g (x) = f (x) ( q + a m b 1 n x m n) g (x) S. It is in S but the degree of this polynomial is m 1, i.e. it is < deg r, contradicting the choice of r as having minimum degree. Hence our assumption is false and deg r < deg g as required. (Uniqueness) Assume for contradiction that, for some f and g, the division is not unique. For this pair we can find distinct divisions f = qg + r and f = q g + r 11

12 with either r 0 or deg r < deg g and either r 0 or deg r < deg g. Then qg + r = q g + r, or r r = (q q) g. Thus g divides r r. This means either r r 0, i.e. r = r, and thus q = q, contradicting the fact that we started with distinct divisions, or deg g deg (r r ) max (deg r, deg r ) < deg g, again a contradiction. unique. Hence our assumption is false and the division is 2) Theorem Let F be as above. Let f, g F [x] be polynomials not both identically zero. Then the gcd (f, g) exists. Proof Let S = {f (x) K (x) + g (x) L (x) : K (x), L (x) Q [x]} {0}. Let D be the set of degrees of polynomials in S. W.l.o.g. assume f is not identically zero and take K 1 and L 0 to see that f = 1 f + 0 g S. Thus S, and and in turn D, are non-empty. D is a set of nonnegative integers so, by the well-ordering principle it has a least element. Take K (x), L (x) F [x] so that d (x) = f (x) K (x) + g (x) L (x) has this minimum degree. We have to show that (i) and (ii) of the definition of gcd both hold for this d (x). (i) Use the division Theorem to write f (x) = q (x) d (x) + r (x) for some q (x), r (x) F [x] with either r 0 or deg r < deg d. Then r (x) = f (x) q (x) d (x) = f (x) q (x) (f (x) F (x) + g (x) L (x)) = f (x) (1 q (x) K (x)) + ( q (x) L (x)) g (x) which is either 0 or in S. If it is not identically zero then we have r (x) S and deg r < deg d, which contradicts the minimality of deg d. Hence r 0 and f (x) = q (x) d (x), i.e. d f. Similarly you can show that d g. (ii) Assume that c (x) divides both f (x) and g (x), so f (x) = c (x) u (x) and g (x) = c (x) v (x) for some u (x), v (x) F [x]. Then d (x) = f (x) K (x) + g (x) L (x) = c (x) u (x) K (x) + c (x) v (x) L (x) = c (x) (u (x) K (x) + v (x) L (x)) and so c (x) divides d (x). 12

13 3) Recall, in Z we chose an m Z, and looked at the congruence classes [a] = {n Z : m (n a)}. We then defined addition and multiplication on the set of classes, Z m. We have seen above many similarities between Z and F [x] where F is one of Q, R, C or Z p, with p prime. So we can try to copy what we did in constructing the congruence classes of Z. Fix a polynomial f (x) F [x] and look at the classes [g (x)] = {h (x) : f (x) (h (x) g (x))}. Then, using the factor notation from the section on relations (since we could define a relation h (x) g (x) if and only if f (x) (h (x) g (x))), we denote the collection of classes by F [x] / (f (x)). We then define addition and multiplication on this set of classes by [g 1 (x)] + [g 2 (x)] = [g 1 (x) + g 2 (x)] and [g 1 (x)] [g 2 (x)] = [g 1 (x) g 2 (x)]. remember, you should check that these are well-defined operations. Example Let f (x) = x R [x]. Given any polynomial p (x) we can use the division theorem to write p (x) = q (x) (x 2 + 1) + r (x) with either r 0 or deg r (x) < deg (x 2 + 2) = 2. Hence [p (x)] = [r (x)] = [a + bx] with a, b R. So all the classes in R [x] / (x 2 + 1) can be labelled simply as [a + bx], a, b R. For addition and multiplication we have [a + bx] + [c + dx] = [(a + c) + (b + d) x] [a + bx] [c + dx] = [ ac + bcx + adx + bdx 2] = [ (ac bd) + (bc + ad) x + db ( x 2 1 )] Note that = [(ac bd) + (bc + ad) x]. (a + bi) (c + di) = (ac bd) + (bc + ad) i. So the map R [x] / (x 2 + 1) C, [a + bx] a+bi is a map that preserves addition and multiplication. Or we can say that R [x] / (x 2 + 1) is a way of constructing the complex numbers from the reals. Example Take x 2 +x+2 Z 3 [x]. Again the classes are of the form [a + bx], a, b Z 3. There are 9 classes, 3 choices for a and 3 for b. They are [0], [1], [2], [x], [1 + x], [2 + x], [2x], [1 + 2x] and [2 + 2x]. 13

14 For the addition and multiplication we can write out the tables table[1].pdf For the multiplication table we need only look at non-zero classes. [1] [2] [x] [1 + x] [2 + x] [2x] [1 + 2x] [2 + 2x] [1] [1] [2] [x] [1 + x] [2 + x] [2x] [1 + 2x] [2 + 2x] [2] [2] [1] [2x] [2 + 2x] [1 + 2x] [x] [2 + x] [1 + x] [x] [x] [2x] [1 + 2x] [1] [1 + x] [2 + x] [2 + 2x] [2] [1 + x] [1 + x] [2 + 2x] [1] [2 + x] [2x] [2] [x] [1 + 2x] [2 + x] [2 + x] [1 + 2x] [1 + x] [2x] [2] [2 + 2x] [1] [x] [2x] [2x] [x] [2 + x] [2] [2 + 2x] [1 + 2x] [1 + x] [1] [1 + 2x] [1 + 2x] [2 + x] [2 + 2x] [x] [1] [1 + x] [2] [2x] [2 + 2x] [2 + 2x] [1 + x] [2] [1 + 2x] [x] [1] [2x] [2 + x] Note that every row (and column) is simply a permutation of the classes, none is missed and none replicated. We see no zero classes, i.e. we have no 14

15 divisors of zero. And all classes have inverses, namely [1] 1 = [1], [2] 1 = [1], [x] 1 = [1 + x], [1 + x] 1 = [x], [2 + x] 1 = [1 + 2x], [2x] 1 = [2 + 2x], [1 + 2x] 1 = [2 + x] and [2 + 2x] 1 = [2x]. Example Student to take x 3 + x + 1 Z 2 [x]. Draw up the tables for (Z 2 [x] / (x 3 + x + 1), +) and (Z 2 [x] / (x 3 + x + 1) \ {[0]}, ). The different classes are [0], [1], [x], [1 + x], [ x 2], [ 1 + x 2], [ x + x 2] and [ 1 + x + x 2]. [1] [x] [1+x] [x 2 ] [1+x 2 ] [x+x 2 ] [1+x+x 2 ] [1] [1] [x] [1+x] [x 2 ] [1+x 2 ] [x+x 2 ] [1+x+x 2 ] [x] [x] [x 2 ] [x+x 2 ] [1+x] [1] [1+x+x 2 ] [1+x 2 ] [1+x] [1+x] [x+x 2 ] [1+x 2 ] [1+x+x 2 ] [x 2 ] [1] [x] [x 2 ] [x 2 ] [1+x] [1+x+x 2 ] [x+x 2 ] [x] [1+x 2 ] [1] [1+x 2 ] [1+x 2 ] [1] [x 2 ] [x] [1+x+x 2 ] [1+x] [x+x 2 ] [x+x 2 ] [x+x 2 ] [1+x+x 2 ] [1] [1+x 2 ] [1+x] [x] [x 2 ] [1+x+x 2 ] [1+x+x 2 ] [1+x 2 ] [x] [1] [x+x 2 ] [x 2 ] [1+x] Note that [x] 2 = [x 2 ], [x] 3 = [1 + x], [x] 4 = [x + x 2 ], [x] 5 = [1 + x + x 2 ], [x] 6 = [1 + x 2 ] and [x] 7 = [1]. 15

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