Polynomials - Olympiad Level MCQ, Class 10 Mathematics

We have a 3 degree polynomial here of form
ax³ + bx² + cx +d = 0
Polynomial p(x) = 2x³ – 6x² – 4x + 30

The roots of this polynomial are α, β, γ
We have formula for products of zeroes taking 2 at a time = αβ + βγ + αγ = c/a
αβ + βγ + αγ = -4/2 = -2

Sum of zeroes = 3

Product of zeroes = -2

Let the quadratic polynomial be ax² + bx + c

Sum of zeroes = 3

-b/a = 3/1

So a = 1, b = -3

Product of zeroes = -2

c/a = -2/1

So a = 1, c = -2

Hence the  quadratic polynomial will be   x² - 3x - 2

Given ,g(x)=x+2=0
x=-2

P(x)=x³-2ax²+16
p(-2)=(-2³) -2 × a × (-2²)+16
=-8 - 8a +16
=8 - 8a
=>-8a = -8
=>a=1

f(x)=ax²+bx+c

It is given that f(0)=5, f(−1)=12 and f(2)=−3

f(0)=c=5 ... (i)

f(−1)=a−b+c=12

∴a−b=12−c

∵c=5

∴a−b=12−5 =7 ... (ii)

f(2)=4a+2b+c=−3

⇒4a+2b=−3−c

as c=5

⇒4a+2b=−8 

⇒2a+b=−4 ... (iii)

Solving (ii) and (iii), we get

a=1

b=−6

Hence, a=1,b=−6,c=5

Thus, a+b+c =1−6+5 =0.

The given polynomials are

f(x) = x² + px + 1

g(x) = x² + qx + 1

Since a, b are the zeroes of f(x),

a + b = - p ..... (1)

ab = 1 ..... (2)

Since c, d are the zeroes of g(x),

c + d = - q ..... (3)

cd = 1 ..... (4)

Now, E = (a - c) (b - c) (a + d) (b + d)

= {ab - (a + b)c + c²} {ab + (a + b)d + d²}

= (1 + pc + c²) (1 - pd + d²) [by (1) & (2)]

= 1 - pd + d² + pc - p²cd + pcd² + c² - pc²d + c²d²

= 1 - pd + d² + pc - p²(cd) + p(cd)d + c² - pc(cd) + (cd)²

= 1 - pd + d² + pc - p² + pd + c² - pc + 1 [by (4)]

= 2 + c² + d² - p²

= 2 + (c + d)² - 2(cd) - p²

= 2 + (- q)² - 2 - p² [by (3) & (4)]

= q² - p²

Since a and b are roots of  x² – px + r = 0 
We can say a+b=p....say equation 1 and ab=r
Also a/2 and 2b are roots of  x² – qx + r  
So, a/2 +2b=q ..... say equation 2 and a/2*2b=ab=r
2(equation1)-equation 2 = 2a -a/2 +2b-2b = 2p-q
so 3a/2=2p-q
a=2(2p-q)/3
Substituting in equation 1
We get b=p-4p/3+2q/3=(2q-p)/3
So we know ab=r=2(2p-q)(2q-p)/9

When x²⁰⁰ + 1 is divided by x² + 1, the remainder is equal to

x²⁰⁰ + 1 =  x²⁰⁰ + (x¹⁹⁸ - x¹⁹⁸ - x¹⁹⁶ + x¹⁹⁶ + x¹⁹⁴  - x¹⁹⁴ - x¹⁹² + x¹⁹² +x¹⁹⁰ - x¹⁹⁰ +..................-x⁴ + x⁴ + x² - x² - 1 + 1 )+ 1

=> x²⁰⁰ + 1  = x²⁰⁰ + x¹⁹⁸ - x¹⁹⁸ - x¹⁹⁶ + x¹⁹⁶ + x¹⁹⁴  - x¹⁹⁴ - x¹⁹² + x¹⁹² +x¹⁹⁰ - x¹⁹⁰ +..................-x⁴ + x⁴ + x² - x² - 1 + 1 + 1

=> x²⁰⁰ + 1  = x¹⁹⁸(x² + 1) - x¹⁹⁶(x² + 1) + x¹⁹⁴(x² + 1)  - x¹⁹²(x² +1) + x¹⁹⁰(x² +1) - x¹⁸⁸(x²+1) +..................+ x²(x² + 1) - (x² + 1) + 1 + 1

=>  x²⁰⁰ + 1  = x¹⁹⁸(x² + 1) - x¹⁹⁶(x² + 1) + x¹⁹⁴(x² + 1)  - x¹⁹²(x² +1) + x¹⁹⁰(x² +1) - x¹⁸⁸(x²+1) +..................+ x²(x² + 1) - 1.(x² + 1) + 2

=>  x²⁰⁰ + 1  = (x² + 1)(x¹⁹⁸- x¹⁹⁶ + x¹⁹⁴ - x¹⁹² + x¹⁹⁰ - x¹⁸⁸+..................+ x² - 1) + 2

=> when x²⁰⁰ + 1 is divided by x² + 1  the reminder = 2

A*(p+q)^2 + 2bpq + c = 0 and a*(p+r)^2 +2bpr + c = 0 
=> aq^2 + 2p(a+b)q + ap^2 + c = 0
and ar^2 + 2p(a+b)r + ap^2 + c = 0 

=> q and r are the roots of the equation 
ax^2 + 2p(a+b)x + ap^2 + c = 0 
=> qr = (ap^2 + c) / a.

(α+β+γ)²=α²+β²+γ²+2(αβ+βγ+αγ)

⇒4=6+2(αβ+βγ+αγ)

⇒αβ+βγ+αγ=−1

(α³+β³+γ³−3αβγ)=(α+β+γ)(α²+β²+γ²−αβ−βγ−αγ)

⇒8−3αβγ=2(6+1)

⇒αβγ=−2

(α²+β²+γ²)²=α⁴+β⁴+γ⁴+2(α²β²+β²γ²+α²γ²)

⇒α⁴+β³+γ⁴+2[(αβ+βγ+αγ)²−2αβγ(α+β+γ)]

⇒36=α⁴+β⁴+γ⁴+2[(−1)²−2(−2(2)]

⇒α⁴+β⁴+γ⁴=36−18

⇒18

Hence (C) is the correct answer

If α,β are the roots of the polynomial x² - px + q.

α + β = p and αβ = q

if the roots are (α² -β²)(α³ - β³) and (α³β² + α²β³)

= (α² -β²)(α³ - β³) + (α³β² + α²β³)

= (α +β)(α -β)2(α² + αβ+ β²) + α²β²(α + β).

= (α +β)[ (α +β)² - 4αβ ] [ (α +β)² - αβ ] + α²β²(α + β).

= p( p² - 4q)( p² - q) + q²p

= p( p4 - p²q -4p²q + 4q²) + pq²

= p⁵ - 5p³q + 5pq² .

⇒ (α² -β²)(α³ - β³)(α³β² + α²β³) = (p⁶q²-5p⁴q³+4p²q⁴)

∴ x²- (p⁵ - 5p³q + 5pq² ) x + ( p⁶q² - 5p⁴q³ + 4p²q⁴ ) = 0

α+β=−b
αβ=ac
Similarly,
α+γ=−a
αγ=bc
Now α is the common root.
Therefore, from the equations,
αβ=ac and αγ=bc
We get the common root α=c
Therefore
β=a and γ=b

Since α+β=−b
Hence
c+a=−b
a+b+c=0

if α and β are two roots of equation ax²+bx+c=0 

product of roots α*β = c/a

now if one root is negative and other positive, then product of the roots should be negative.

i.e c/a is negative which means both a and c are of opposite signs.

sum of roots α+β= -b/a = p

product of roots αβ = c/a= 36

α²+β²= 9

α²+β² +2αβ - 2αβ = 9

(α+β)² - 2αβ = 9

p²= 9 + 2*36

p² = 81

p = 9

α+β = -b/a    (eq1)    αβ = c/a  (eq 2)

let the second equation cx² + bx + a has roots x and y

x+y = -b/c   (eq 3)   xy = a/c (eq 4)

divide eq 1 by eq 3 

α+β / x+y = (-b/a) / (-b/c)

                = c/a

                = αβ

α+β / αβ = x+y

x+y = 1/α + 1/β

a transcendental number is a complex number that is not algebraic—that is, not a root (i.e., solution) of a nonzero polynomial equation with integer or equivalently rational coefficients. The most popular transcendental numbers are π and e

As the roots are 1,2,4/3,-1

Therefore the equation will be

(x-1)(x-2)(x-4/3)(x+1)

If we will solve this thing then we will get

3x⁴ – 10x³ + 5x² + 10x – 8

roots are 4,3/2 and – 2 so we can form equation as

(x-4)(x-3/2)(x-2)=0

which will give 2x³ – 7x² – 10x + 24

Let α,β,γ be the zeroes of given polynomial p(x)......
then,
α+β+γ= -b/a = 2   (given)
=> 2 = -(-5)/k
=> k =5/2

for a cubic equation product of roots is α,β,γ = -d/a

    = -(-1)/4   = 1/4

for a equation of 4 roots ax⁴+bx³+cx²+dx+e=0 

sum of roots is -b/a

in our equation b = 0 (since there is no term of x³)

so sum of roots is -b/a=0

x² – ax + b = 0 and x² – px + q = 0 are the equations 

second equation has two equal roots let the roots be k

2k = p => k = p/2   and k²=q 

now both equations has one common root k 

for first equation k+ l = a => l = a- p/2      and kl = b

put k and l

p/2 (a-p/2) = b 

ap/2 - p²/4 = b

ap/2 = k²+b

ap/2 = q+b