Polynomials Test - 15

It is known that: (a − b − c)² = a² + b² + c² − 2ab + 2bc − 2ac

∴(a − b − c)² = a² + b² + c² + 2 (bc − ac − ab)

⇒ a² + b² + c² = (a − b − c)² − 2 (bc − ac − ab)

= 49 − 2(12) [(a − b − c)² = 49, bc − ac − ab = 12]

= 25

The correct answer is B.

The given polynomial can be factorised as:

125x³ − 64y³ − 300x²y + 240xy²

= (5x)³ − (4y)³ − 3 (5x)² (4y) + 3 (5x) (4y)²

= (5x)³ − (4y)³ − 3 (5x) (4y) (5x − 4y)

= (5x − 4y)³ [(a − b)³ = a³ − b³ − 3ab (a − b)]

= (5x − 4y) (5x − 4y) (5x − 4y)

The correct answer is C.

It is known that: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Putting x = a, y = −2b, and z = −3c:

(a − 2b − 3c)² = (a)² + (−2b)² + (−3c)² + 2(a) (−2b) + 2 (−2b) (−3c) + 2(a) (−3c)

= a² + 4b² + 9c² − 4ab + 12bc − 6ac

The correct answer is B.

It is given that (x − 1) is a factor of the polynomial x³ + kx² − 8x.

Therefore, x = 1 is a zero of the polynomial x³ + kx² − 8x.

It is known that the value of a polynomial at its zero is zero.

∴ p (1) = 0

Thus, the value of p (1) is zero.

The correct answer is B.