Good Challenge

Algebra Level 2

The quadratic polynomial a x 2 + b x + c ax^2 + bx + c , when divided by the linear factors x , x − 1 x, x-1 and x − 2 , x-2, leaves remainder 1, 2, and 9 respectively.

What is the value of a + b + c a + b + c ?

3 2 5 4

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Jayesh Agrawal by Jayesh Agrawal , 32, India

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3 solutions

Boris Barron
Aug 25, 2014

Using long division of polynomials:

a x 2 + b x + c x − 1 = a x + b + a + a + b + c x − 1 \frac{ax^2+bx+c}{x-1} = ax+ b + a + \frac{a+b+c}{x-1}

Remainder is a + b + c a + b + c , so a + b + c = 2 a + b + c = 2

interestingly, is not necessary to divide by x x or x − 2 x-2 .

17 Helpful 0 Interesting 0 Brilliant 0 Confused

Smart one :-).

Bhargav Upadhyay - 6 years, 4 months ago

U s i n g t h e s e c o n d g i v e n c o n d i t i o n , L e t E q u a t i o n p ( x ) w h e n d i v i d e d b y ( x − 1 ) g i v e s r e m a i n d e r 2. S o p u t x = 1 i n p ( x ) , ⇒ a ( 1 ) 2 + b ( 1 ) + c = 2 ⇒ a + b + c = 2 \quad \quad Using\quad the\quad second\quad given\quad condition,\quad Let\quad Equation\quad p(x)\\ \quad \quad when\quad divided\quad by\quad (x-1)\quad gives\quad remainder\quad 2.\\ \quad \quad So\quad put\quad x\quad =\quad 1\quad in\quad p(x),\\ \Rightarrow a(1)^{ 2 }+b(1)+c\quad =\quad 2\\ \Rightarrow a+b+c\quad =\quad 2

CHEERS:)

14 Helpful 0 Interesting 0 Brilliant 0 Confused

I pressed the wrong button again. Old age.

This is a good one for applying the Python sympy library. First we divide the general quadratic with each of the binomial factors, requiring division in the ring of polynomials with integer coefficients (that's the domain thing), to obtain expressions for the required remainders.

>>> from sympy import *

>>> a,b,c,x=symbols('a b c x')

>>> div(a x 2+b x+c,x,domain='ZZ')

(a*x + b, c)

>>> div(a x 2+b x+c,x-1,domain='ZZ')

(a*x + a + b, a + b + c)

>>> div(a x 2+b x+c,x-2,domain='ZZ')

(a x + 2 a + b, 4 a + 2 b + c)

Then we equate these remainders with the required numerical values and solve:

4 a + 2 b + c = 9 a + b + c = 2 c = 1 4a+2b+c=9 a+b+c=2 c=1

to get

a = 3 , b = − 2 a n d c = 1 a = 3, b = -2 and c = 1

Now we can verify that we have the correct values:

>>> div(3 x 2-2 x+1,x)

(3*x - 2, 1)

>>> div(3 x 2-2 x+1,x-1)

(3*x + 1, 2)

>>> div(3 x 2-2 x+1,x-2)

(3*x + 4, 9)

Bill Bell - 6 years, 9 months ago

p(x)=q(x)(x-1)+2 so p(1)=2=a+b+c

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