Question

49. Find the quadratic polynomial when divided by
x, x – 1 and x - 2 leaves remainders 1, 2 and 9
respectively.​

Answer 1

I’m sure this isn’t the most elegant solution, but I think it’s quite easy to follow.

Let our quadratic expression be  ax2+bx+c  

Now,  ax2+bx+c=(ax+b)x+c  

The first part is clearly divisible by x, leaving the remainder c. Thus c = 1.

 Now,  x=(x−1)+1⇒x2=[(x−1)+1]2  

=(x−1)2+2(x−1)+1  

We can thus rewrite the expression as:

a(x−1)2+2a(x−1)+a+b(x−1)+b+1  

=a(x−1)2+(2a+b)(x−1)+(a+b+1)  

The first two parts are clearly divisible by (x-1), leaving the remainder a + b + 1 = 2. This implies that a + b = 1.

Also,  x=(x−2)+2⇒x2=[(x−2)+2]2  

=(x−2)2+4(x−2)+4  

We can thus rewrite the expression as:

a(x−2)2+4a(x−2)+4a+b(x−2)+2b+1  

=a(x−2)2+(4a+b)(x−2)+(4a+2b+1)  

The first two parts are clearly divisible by (x-2), leaving the remainder 4a + 2b + 1 = 9. This implies that 4a + 2b = 8, thus 2a + b = 4.

We now have a pair of simultaneous equations:

a + b = 1

2a + b = 4

Subtracting Equation 1 from Equation 2, we have a = 3, which means that b = -2.

Answer:  3x2−2x+1