Question

The polynomials x3 + ax2 + 9x + 6 and x3 + bx2 + 6x + 3 have a common quadratic factor over the set of polynomials with integral coefficients. Find the value of (a + b).

Answer 1

Correct Question:

The polynomials x3 + ax2 + 9x + 6 and x3 + bx2 + 6x + 3 have a common quadratic factor over the set of polynomials with integral coefficients. Find the value of (a + b).

Solution :

$let \: x {}^{3} + ax {}^{2} + 11x + 6 = m.....(i) \\ and \: x {}^{3} + bx {}^{2} + 14x + 8 = n......(ii) \\ \: subtract \: 1 \: from \: 2 \\ ( b - a) + 3x + 2 = (n - m)......(iii) \\ x {}^{2} + px + q = u...(4)$

Comparing 3 and 4

we get b - a =1 = b = a + 1, p =3 and q = 2

Now, b + q = (a +1) +2 = a + 3 = a +p

So a +p = b + q

Hence, proved

Thanks!

Answer 2

Answer:

(a,b)=(5,4)

Step-by-step explanation:

Let the required quadratic factor be x2 + px + q

This is also a factor of difference of given polynomials.

⇒ (x2 + px + q) N = (a – b)x2 + 3x + 3 for some integer N.

Since p and q are integers, N must divide 3,

So, N = 1 or N = 3

If N = 1, a – b = 1 as well.

Suppose x3 + bx2 + 6x + 3 = (x2 + 3x + 3) (x + r) for some integer r.

Clearly, r = 1, and multiplication produces

b = 4, so a = 5.

We can check that x3 + 5x2 + 9x + 6 is divisible by x2 + 3x + 3, which it is.

If N = 3, the common quadratic factor would be

x2 + x + 1, and we would need

x3 + bx2 + 6x + 3 = (x2 + x + 1) (x + s) for some integral s.

Clearly s = 3, but then (x2 + x + 1) (x + 3) does not produce 6x term.

∴ only x2 + 3x + 3 works.

∴ (a, b) = (5, 4)