Answer :

Given:

HCF = x – 1

Let the two polynomials be u(x) and v(x)

So (x – 1) is common to both u(x) and v(x)

LCM = x^{3} – 7x + 6

= x^{3} – 7x + 6

= x^{3} – (1 + 6) x + 6

= x^{3} – x - 6x + 6

= x(x^{2} – 1) - 6(x - 1)

(x^{2} – 1) = (x + 1) (x – 1)

a^{2} – b^{2} = (a + b) (a – b)

Here a = x and b = 1

= x(x + 1) (x – 1) - 6(x - 1)

= (x-1) [x(x + 1) - 6]

Now solving the inner quadratic equation,

x^{2} + x - 6 = 0

x^{2} + 3x - 2x - 6 = 0

x(x + 3) - 2(x + 3) = 0

(x + 3) (x – 2) = 0

∴ LCM = (x-1) (x + 3) (x – 2)

Since HCF = (x-1), which implies that both u(x) and v(x) contains (x – 1)

Therefore u(x) = (x – 1) (x + 3)

= x^{2} + 2x - 3

v (x) = (x – 1) (x – 2)

= x^{2} – 3x + 2

Therefore the polynomial are x^{2} + 2x – 3 and x^{2} – 3x + 2.

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