Download MATHEMATICS ENRICHMENT - POLYNOMIALS Q1. Find all

MATHEMATICS ENRICHMENT - POLYNOMIALS
Q1. Find all integers n for which the following expressions are perfect squares:
a) n2 + 20n + 11
b) n2 + 20n + 12
c) n2 − 11n + 63
Hint: Complete the square (n + y)2 where y = half the coefficient of n. Then use
the difference of squares formula.
Q2. Show that for any integer n > 11, the expression n2 − 19n + 89 is never
a perfect square.
Hint: Try the same method as in Q1, prove that your solutions are not
integers.
Q3. Show that for any non-negative integer n, that 3n + 2(17n ) is never a
perfect square.
Note: This problem is not really about polynomials so its solution will come
by different methods. For example, taking n = 1, 2, 3, 4, ... you can systematically
check the last digit of 3n + 2(17n ). Can a perfect square end in those digits?
Q4. Simplify the following
a) (a − b)3 + (b − c)3 + (c − a)3
b) (a + 2b − 3c)3 + (b + 2c − 3a)3 + (c + 2a − 3b)3
Hint: a) Method I: Use X 3 + Y 3 = (X + Y )(X 2 − XY + X) for the sum
(a − b)3 + (b − c)3 . After simplifying, can you find a common factor for all 3 terms
in the sum at a)? Method II: Think of this as a polynomial in the variable a like
this:
P (a) = (a − b)3 + (b − c)3 + (c − a)3 .
Think of b and c as numbers. Sub in values for a to find how to factor the
polynomial.
Q5. Show that (a + b + c)3 − a3 − b3 − c3 = 3(a + b)(b + c)(c + a).
Hint: a) Method I: Multiply through. Method II: Think of these as polynomials in the variable a and find its factors. Also use the symmetry in a, b, c.
Q6. Factorise the following expressions
a) x4 + x2 + 1.
a) Hint: Add and subtract x2 and use differences of squares.
5
b) x + x + 1
b) Hint: Add and subtract x2 and use x3 − 1 = (x − 1)(x2 + x + 1).
9
c) x + x4 − x − 1
1
2
c) Hint: Group the 4 terms of the sum in pairs of 2 and use various
factorization formulae like for example
x5 + 1 = (x + 1)(x4 − x3 + x2 − x + 1) or
x8 − 1 = (x4 − 1)(x4 + 1) = (x2 − 1)(x2 + 1)(x4 + 1) = (x − 1)(x + 1)(x2 + 1)(x4 + 1)
d) x12 − x4 − x3 + 1
d) Hint: Similar to the above
e) x10 + x5 + 1
e) Hint: What is the remainder when dividing by x3 − 1?
Q7. Find all positive integers, n, for which n8 + n + 1 is prime. Q8. Prove or
disprove the following statement: For every positive integer, n, the greatest common divisor of 5n + 4 and 9n - 7 is 1. Q9. Show that if x is a non-zero real number,
then x8 - x5 −1x+1x4 0Q10.Letn > 0beaninteger.F indtheremainderupondivisionof xn+
xn − 1 + +1by(a)x2 + 1and(b)x2 + x + 1.12M AT HEM AT ICSEN RICHM EN T −
P OLY N OM IALSQ11.F indthegreatestcommondivisorof thef ollowingpairsof polynomialsf (x) =
x2 −1, g(x) = x3 −1f (x) = x2 +x, g(x) = x3 −x2 +xf (x) = x3 −3x−2, g(x) = x3 +
x2 −3x−6f (x) = x3 +3x2 +5x+3, g(x) = x4 +x3 +x2 +2x+1f (x) = x4 −16, g(x) =
x4 +8x3 +24x2 +32x+16f (x) = x5 +x4 +x3 +x2 +x+1, g(x) = x3 +x2 +x+1f (x) =
x6 + x4 + 2x + 1, g(x) = x7 − x5 + x4 + 3x3 − 2x2 + 2Q12.F orthef ollowing
nd the greatest common divisor and write it as a linear combination of these
two polynomials f(x) = x3 − 3x − 2, g(x) = x3 + x2 − 3x − 6f (x) = x3 − 1, g(x) =
x4 −x3 −x2 +1f (x) = x3 +1, g(x) = x5 +1Q13.Letn > 1beaninteger.P rovethatn5+
n + 1isdivisiblebyatleasttwodistinctprimes.Q14.Showthatif ndividesm, thenxn −
1dividesxm−1.Q15.Bytestingaf ewvalues, guessthevalueof gcd(xn−1; xm−1)f orarbitraryn; m >
0.Q16.F indallpositiveintegersnsuchthatx2 n+1+x+1isdivisiblebyxn+x+1.Q17.Giventhatapolynomialf (x
g(x), andthedierenceof thepolynomials, g(x)−f (x).W iththisf actinmind, tryapproachQ16inadierentmann
1. Find all solutions to the following equations:
a)x3 − 5x2 + 4 = 0.
Hint: Writing x3 − 5x2 + 4 = (x − a)(x − b)(x − c) and then multiplying
through the right hand side and identifying the coefficients we get −abc = 4. Thus
if a, b or c is an integer, then it must be a divisor of 4.
b)x3 + 2x2 − x + 6.
Hint: Integer solutions must be divisors of 6.
c)x3 − 6x2 + 12x − 10 = 0.
Hint: First complete the cube x3 − 6x2 + 12x + 8 = (x − 2)3 .
2. Find the sums Sn = r1n + r2n + r3n for n = 1, 2, ...5 if r1 , r2 and r3 are the
roots of the equation
x3 − 2x2 + 2x + 3 = 0.
3
Hint: Writing x3 − 2x2 + 2x + 3 = (x − r1 )(x − r2 )(x − r3 ) and multiplying
the right hand side we get r1 + r2 + r3 = 2 as well as r1 r2 + r1 r3 + r2 r3 = 2 and
r1 r2 r3 = −3.
Then check r12 + r22 + r32 = (r1 + r2 + r3 )2 − 2(r1 r2 + r1 r3 + r2 r3 ) and use this to
find S2 .
To find S3 , note that substituting x = ri in the equation
x3 − 2x2 + 2x + 3
=⇒ r13 − 2r12 + 2r1 + 3
r23 − 2r22 + 2r2 + 3
r33 − 2r32 + 2r3 + 3
=
=
=
=
(x − r1 )(x − r2 )(x − r3 )
0.
0.
0.
Add the last three equations to find S3 − 2S2 + 2S1 + 9 = 0 and hence find S3 .
To find Sn , first multiply the equations above by the suitable rin−3 and then add
them up to find Sn − 2Sn−1 + 2Sn−2 + 3Sn−3 = 0.
3. Find the numbers a and b knowing that if you subtract the number 7
from each of the roots of the equation x2 + ax + b = 0, you will obtain the roots
of x2 + bx + a = 0.
Hint: Write x2 + ax + b = (x − r1 )(x − r2 ), multiply through and compare
to (x − r1 + 7)(x − r2 + 7) = x2 + bx + a. 4. A certain polynomial p(x) has the
values p(2) = 4 and p(−1) = 5. If you divide p(x) by x2 − x − 2, what remainder
will you get? (First decide what will be the degree of the remainder).
Hint: Suppose that long division of p(x) by x2 − x − 2 gives us q(x) together
with a remainder r(x). Note that we can keep dividing by x2 − x − 2 until we get
r(x) of degree 1, namely of the form r(x) = Ax + B.
p(x) = (x2 − x − 2)q(x) + Ax + B.
Substitute x = 2 and x = −1.
5. Let p(x) be a polynomial of degree ≤ 2 and a, b, c be some numbers.
Prove that
p(a + b + c) − p(a + b) − p(a + c) − p(b + c) + p(a) + p(b) + p(c) − p(0) = 0
.
First check that the equation holds for any p(x) = xn .