TS ICET 23 May 2018 Shift 2 Paper

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Question : 123 of 200

Marks: +1, -0

For two polynomials p(x) and q(x) the highest common factor is (x - 2) and the least common multiple is

\x^3-9\x^2+26\x - 24.

If p(x) =

\x^2 -6\x+8

then q(x) =

$\x^2 +5\x + 6$
$\x^2 -5\x - 6$
$\x^2 -5\x + 6$
$\x^2 -2\x + 3$

Solution:

Given

p(x) = x^2 – 6x + 8

=(x-2)(x-4)

Let

q(x) = (x – 2) (x – k)

Given LCM of

p(x), q(x)=x^{3}-9 x^{2}+26 x-24

∴ (x-2)(x-4)(x-k)=x^{3}-9 x^{2}+26 x-24

⇒ x^{3}-(6+k) x^{2}+(8+6 k) x-8 k=x^{3}-9 x^{2}+26 x-24

∴ Comparing

x^2

coefficients

6+k=9 ⇒ k = 3

∴ q(x)=(x-2)(x-3)=x^{2}-5 x+6

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