The values of a, b and c respectively for the expression f(x) = x3 + a

1.	The values of a, b and c respectively for the expression f(x) = x3 + ax2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are :(a) 9, 20, 12 (b) –9, –20, 12 (c) –9, 20, –12 (d) –9, –20, –12
Answer» (c) –9, 20, –12 Given, f(x) = x³ + lx² + mx + n. f(1) = f(2) = 0 ⇒ (x – 1) and (x – 2) are factors of f(x). Since, f(x) is polynomial of degree 3, it shall have three linear factors. So, let the third factor be (x – k). Then, f(x) = (x – 1) (x – 2) (x – k) ⇒ f(x) = x³ + lx² + mx + n = (x – 1) (x – 2) (x – k) Given, f(4) = f(0) ⇒ (4 – 1) (4 – 2) (4 – k) = (–1) (–2) (–k) ⇒ 24 – 6k = – 2k ⇒ 4k = 24 ⇒ k = 6 ∴ f(x) = (x – 1) (x – 2) (x – 6) = (x² – 3x + 2) (x – 6) = x³ – 9x² + 20x – 12 ∴ x³ + lx² + mx + n = x³ – 9x² + 20x – 12 ⇒ l = – 9, m = 20, n = – 12.

The values of a, b and c respectively for the expression f(x) = x3 + ax2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are :(a) 9, 20, 12 (b) –9, –20, 12 (c) –9, 20, –12 (d) –9, –20, –12

Answer»

(c) –9, 20, –12

Given, f(x) = x³ + lx² + mx + n.

f(1) = f(2) = 0 ⇒ (x – 1) and (x – 2) are factors of f(x).

Since, f(x) is polynomial of degree 3, it shall have three linear factors. So, let the third factor be (x – k).

Then, f(x) = (x – 1) (x – 2) (x – k)

⇒ f(x) = x³ + lx² + mx + n = (x – 1) (x – 2) (x – k)

Given, f(4) = f(0)

⇒ (4 – 1) (4 – 2) (4 – k) = (–1) (–2) (–k)

⇒ 24 – 6k = – 2k ⇒ 4k = 24 ⇒ k = 6

∴ f(x) = (x – 1) (x – 2) (x – 6) = (x² – 3x + 2) (x – 6)

= x³ – 9x² + 20x – 12

∴ x³ + lx² + mx + n = x³ – 9x² + 20x – 12

⇒ l = – 9, m = 20, n = – 12.