Question

Q1. Find the zero of polynomial...P(x)=x(x-2)(x+3)
Q2. If the polynomial 2xpower 3 +ax power2 +3x -5 and x power 3+x power 2 -4x+a leave the same remainder when divided by x-2 .....find the value of a

Answer 1

(1). p(x) = x(x-2)(x+3)
x(x-2)(x+3) = 0
if x = 0 then x = 0
if x-2 = 0 then x = 2
if x+3 = 0 then x = -3
therefore the zeros of the given polynomial are 0, 2 and -3.
(2). the given polynomial are
$2 {x}^{3} + a {x}^{2} + 3x - 5......(1) \\ {x}^{3} + {x}^{2} - 4x + a ......(2)$
when divided by x-2, remainder is equal in both the cases.
let x-2 = 0
x = 2
by remainder theorem putting x =2 in both the polynomials we get,
$2( {2)}^{3} + a {(2)}^{2} + 3(2) - 5 \\ 2(8) + a(4) + 6 - 5 \\ 16 + 4a + 1 \\ 4a + 17..........(3)$
and
${x}^{3} + {x}^{2} - 4x + a \\ {(2)}^{3} + {(2)}^{2} - 4(2) + a \\ 8 + 4 - 8 + a \\ 4 + a........(4)$
since the remainders are equal therefore (3)=(4)
$4a + 17 = 4 + a \\ 3a = - 13 \\ a = - \frac{13}{3}$

hope this helps you