Tomato Objective 288 | Finding big remainder in a small way

Try this problem from TOMATO Objective 288, useful for ISI BStat, BMath Entrance Exam based on finding big remainder in a small way.

Problem: Tomato objective 288

The remainder R(x) obtained by dividing the polynomial [latex]x^{100}[/latex] by the polynomial [latex]x^2-3x+2[/latex] is

(A) [latex]2^{100}-1[/latex]

(B) [latex](2^{100}-1)x-(2^{99}-1)[/latex]

(C) [latex]2^{100}x-3(2^{100})[/latex]

(D) [latex](2^{100}-1)x+(2^{99}-1)[/latex]

SOLUTION: (B)

The the divisor is a quadratic term .So, R(x) must be 1 degree less than divisor.

We know , [latex]f(x)=divisor.Q(x)+R(x)[/latex]

[latex]=> x^{100}=(x^2-3x+2).Q(x)+ (ax+b)[/latex]

[latex]=> x^{100}=(x-1)(x-2).Q(x)+ (ax+b)[/latex]

when ,[latex]x=2[/latex]

[latex]=> 2^{100}=(2-1)(2-2).Q(x)+ (2a+b)[/latex]

[latex]=> 2^{100}= (2a+b)..........(i)[/latex]

when ,[latex]x=1[/latex]

[latex]=> 1^{100}=(1-1)(1-2).Q(x)+ (a+b)[/latex]

[latex]=> 1^{100}=(a+b)...........(ii)[/latex]

solving two equation we get, [latex]a=(2^{100}-1)[/latex]

and, [latex]b=-(2^{99}-1)[/latex]

The remainder R(x) is [latex](2^{100}-1)x-(2^{99}-1)[/latex]

Some Useful Links:

Condition of real roots | Tomato objective 291

Problem: If the roots of the equation ${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$, (where a,b,c are real numbers) are equal , then

(A) $b^2-4ac=0$

(B) $a=b=c$

(C) a+b+c=0

(D) none of foregoing statements is correct

Answer: $(B)$

${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$

=> $x^2-{(a+b)}x$+$ab+x^2-{(b+c)}x$+$bc+x^2-{(c+a)}x+ca$=$0$

=> $3x^2-2{(a+b+c)}x$+$(ab+bc+ca)$=$0$

discriminant, of the equation is

=> $4{(a+b+c)^2}$-$4.3{(ab+bc+ca)}$=$0$

=> $a^2+b^2+c^2+2(ab+bc+ca)$-$3(ab+bc+ca)$=$0$

=> $a^2+b^2+c^2$-$(ab+bc+ca)$=$0$

=> $a=b=c$

So, option (B) is correct.

Real Roots of a Cubic Polynomial | TOMATO Objective 258

Try this beautiful problem from TOMATO Objective no. 258 based on Real Roots of a Cubic Polynomial.

Problem: Real Roots of a Cubic Polynomial

Let a,b,c be distinct real numbers. Then the number of real solution of [latex](x-a)^3+(x-b)^3+(x-c)^3=0[/latex] is

(A) 1

(B) 2

(D) depends on a,b,c

Solution: Ans: (A)

Let [latex]f(x)=(x-a)^3+(x-b)^3+(x-c)^3[/latex]

[latex]=> f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2=0[/latex]

[latex]=> f'(x)=(x-a)^2+(x-b)^2+(x-c)^2=0[/latex]

[latex]=> x=a, x=b, x=c [/latex]

But it is not possible a quadratic equation has three roots.so, it implies that f'(x) has no real roots.But f(x) is a cubic polynomial. And we know a cubic polynomial must have at least one real root ( we know all polynomial curves are continuous ,so it cuts either of the axes at least once).So, number of real root of the given equation is 1 .

Roots of a Quintic Polynomial | TOMATO Objective 257

Try this beautiful problem from TOMATO Objective no. 257 based on Roots of a Quintic Polynomial.

Problem: Roots of a Quintic Polynomial

The number of real roots of [latex] x^5+2x^3+x^2+2=0[/latex] is

(A) 0

(B) 3

(C) 5

(D) 1

Solution: Answer: (D)

[latex] x^5+2x^3+x^2+2=0[/latex]

[latex] \implies x^3(x^2+2)+(x^2+2)=0[/latex]

[latex] \implies (x^3+1)(x^2+2)=0[/latex]

[latex] \implies (x+1)\bold{\underline{(x^2-x+1)(x^2+2)}}=0[/latex]

The expression in underline doesn't have any real roots.

Therefore, only real root of the equation is [latex] x=-1[/latex]