If x – 2 and x – \(\frac1{2}\)both are the factors of the polynomi

1.	If x – 2 and x – \(\frac1{2}\)both are the factors of the polynomial nx2 – 5x + m, then show that m = n = 2.
Answer» p(x) = nx² – 5x + m (x – 2) is a factor of nx² – 5x + m. ∴ By factor theorem, P(2) = 0 ∴ p(x) = nx² – 5x + m ∴ p(2) = n(2)²– 5(2) + m ∴ 0 = n(4) – 10 + m ∴ 4n – 10 + m = 0 …(i) Also, ( x = \(\frac1{2} \)) is a factor of nx² – 5x + m. ∴ By factor theorem, p( \(\frac1{2} \)) = 0 p(x) = nx²– 5x + m ∴ p(\(\frac1{2} \) ) = n(\(\frac1{2} \))² – 5\(\frac1{2} \) + m 0 = \(\frac{n}{4} \)–\(\frac{5}{2} \) + m ∴ 0 = n- 10 +4m … [Multiplying both sides by 4] ∴ n = 10 – 4m ……(ii) Substituting n = 10 – 4m in equation (i) 4(10 – 4m) – 10 + m = 0 ∴ 40 – 16m – 10 + m = 0 ∴ -15m+ 30 = 0 ∴ -15m = -30 ∴ m = 2 Substituting m = 2 in equation (ii), n = 10 – 4(2) = 10 – 8 ∴ n = 2 ∴ m = n = 2

If x – 2 and x – \(\frac1{2}\)both are the factors of the polynomial nx2 – 5x + m, then show that m = n = 2.

Answer»

p(x) = nx² – 5x + m

(x – 2) is a factor of nx² – 5x + m.

∴ By factor theorem,

P(2) = 0

∴ p(x) = nx² – 5x + m

∴ p(2) = n(2)²– 5(2) + m

∴ 0 = n(4) – 10 + m

∴ 4n – 10 + m = 0 …(i)

Also, ( x = \(\frac1{2} \)) is a factor of nx² – 5x + m.

∴ By factor theorem,

p( \(\frac1{2} \)) = 0

p(x) = nx²– 5x + m

∴ p(\(\frac1{2} \) ) = n(\(\frac1{2} \))² – 5\(\frac1{2} \) + m

0 = \(\frac{n}{4} \)–\(\frac{5}{2} \) + m

∴ 0 = n- 10 +4m … [Multiplying both sides by 4]

∴ n = 10 – 4m ……(ii)

Substituting n = 10 – 4m in equation (i)

4(10 – 4m) – 10 + m = 0

∴ 40 – 16m – 10 + m = 0

∴ -15m+ 30 = 0

∴ -15m = -30

∴ m = 2 Substituting m = 2 in equation (ii),

n = 10 – 4(2)

= 10 – 8

∴ n = 2

∴ m = n = 2