Answer :

Given, p(x) = 6x^{2} – 7x + 2

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ 6x^{2} – 7x + 2 can be written as 6(x^{2} – x + ) = 6(x^{2} – x + )

⇒ Now, we will find out the factors for x^{2} – x +

⇒ x^{2} – x + is in the form x^{2} – (a + b)x + ab and (x – a)(x – b) are the factors

∴ coefficient on either side of the polynomial is same and a + b =

and ab =

⇒ we know that (a + b)^{2} –(a – b)^{2} = 4ab

∴ (a + b)^{2} – 4ab = (a – b)^{2}

⇒ – 4() = (a – b)^{2}

⇒ – = (a – b)^{2}

⇒ = (a – b)^{2}

⇒ = (a – b)^{2}

⇒ a – b =

⇒ a – b =

⇒ We need to find out the values of a and b by solving (a + b) + (a – b) and (a + b) – (a – b)

⇒ we take a – b =

⇒ (a + b) + (a – b) = +

⇒ 2a =

⇒ a = =

⇒ (a + b) – (a – b) = –

⇒ 2b =

⇒ b =

∴ we get a = and b =

And if we take a – b = – we get as follows

⇒ (a + b) + (a – b) = + (– )

⇒ 2a =

⇒ a =

⇒ (a + b) – (a – b) = – (– )

⇒ 2b =

⇒ b =

∴ a = and b =

⇒ So, we have 6(x – )(x – )

⇒ (6x – )( x – )

⇒ (6x – 3)(x – )

∴ (6x – 3)(x – ) are the factors

⇒ p(x) = 0 if (6x – 3) is 0 and (x – ) is 0

⇒ (6x – 3) = 0

⇒ 6x = 3

⇒ x = =

⇒ Substitute, x = in 6x^{2} – 7x + 2 = 6(^{2} – 7( + 2 = – + 2 = 0

⇒ (x – ) = 0

⇒ x = substitute, 6x^{2} – 7x + 2 = 6()^{2} – 7() + 2 = – + 2 = 0

Hence, (6x – 3)(x – ) are the first degree factors of the polynomial and and are the solution of the given polynomial 6x^{2} – 7x + 2

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