Question

If -3/2 is a zero of a polynomial p(X)=2x^3+9x^2-x-a, find the value of 'a'​

Answer 1

given -3/2 is the zero of the polynomial..

it means it satisfies the given polynomial...

if x is the root and p(y)is polynomial ..

then p(x)=0...

now substitute..-3/2 in given polynomial..

p(-3/2)=0

2×(-3/2)^3+9×(-3/2)^2-(-3/2)-a=0

15-a=0

a=15

Answer 2

Answer:

Value of a = 15

Step-by-step explanation:

$Given \: \frac{- 3}{2} \: is \: the \: zero \: of \: the \:polynomial$

$p(x) = {2x}^{3} + {9x}^{2} - x - a$

$Substitute \: \frac{ - 3}{2} \: in \: place \: of \: x$

$p( \frac{ - 3}{2}) = 2( { \frac{ - 3}{2}) }^{3} + 9( { \frac{ - 3}{2}) }^{2} - \frac{ - 3}{2} - a = 0$

$2( \frac{ - 27}{8} ) + 9( \frac{9}{4} ) - (\frac{ - 3}{2} ) - a = 0$

$\frac{ - 54}{8} + \frac{81}{4} + \frac{3}{2} - a = 0$

$\frac{ - 54 + 162 + 12 - 8a}{8} = 0$

$\frac{120 - 8a}{8} = 0$

$120 - 8a = 0 \times 8$

$120 - 8a = 0$

$- 8a = 0 - 120$

$- 8a = - 120$

$8a = 120$

$a = \frac{120}{8}$

$a = 15$

Therefore value of a = 15