Question

If 2p3 - xp2 – yp – 14 has (p + 2) as a factor and leaves a remainder -26 when divided by p + 3. find x and y.
Option
a x = 1, y = -17
b X-1. y = -17
c x=-1.y = 17
d x = 1, y = 17​

Answer 1

Answer:

kindly see the answer below...

Step-by-step explanation:

HERE,

2p³- xp²– yp – 14 has (p + 2) as a factor...

so,

$p + 2 = 0$

$or \: \: p = ( - 2)$

then,

$f(p) = 2 {p}^{3} - x {p}^{2} - yp - 14$

$or \: \: f( - 2) = - 16 - 4x + 2y - 14 = 0$

$or \: \: 2y - 4x = 30$

$or \: \: y - 2x = 15 \: \: \: \: ..... {eq}^{n} 1$

Therefore, leaves a remainder -26 when divided by (p + 3).

so,

$p + 3 = 0$

$or \: \: p = ( - 3)$

then,

$f(p) = 2 {p}^{3} - x {p}^{2} - yp - 14$

$or \: \: f( - 3) = - 54 - 9x + 3y - 14 = - 26$

$or \: \: 3y - 9x = 42 \: \: \: \: \: ..... {eq}^{n} 2$

Now, comparing equation 1 &2...we get..

$x = 1$

and,

$y = 17$

so, the right option is (d)

thanks a lot..