Question

If zero of x^3-ax^2+x+3 is a, find the value of a.

Answer 1

ANSWER:

• The value of a = (-3).

GIVEN:

• P(x) = x³-ax²+x+3
• One zero = 'a'

TO FIND:

• Value of 'a'.

SOLUTION:

P(x) = x³-ax²+x+3

Here 'a' is the zero of the given polynomial then we will put 'a' in place of 'x' and we will get the remainder 0.

Putting x = a we get;

p(a) = a³-a(a²)+a+3

=> a³-a³+a+3 = 0

=> a+3 = 0

=> a = (-3)

NOTE:

some important formulas:

$\implies \: \alpha \: + \beta \: = \frac{ - b}{a} \\ \\ \implies \: \alpha \beta \: = \frac{c}{a}$

Where :

• a = coefficient of x²
• b = coefficient of x
• c = constant term.

Some important identities:

• (a+b)²= a²+b²+2ab
• (a-b)²= a²+b²-2ab

Answer 2

GIVEN:

Cubic polynomial = x³ - ax² + x + 3

One of the zeroes of polynomial is a

TO FIND:

The value of "a"

SOLUTION:

We know that, when we substitute zeroes in the polynomial then the value of polynomial will be '0'.

p(x) = x³ - ax² + x + 3

p(a) = (a)³ + a(a)² + (a) + 3

a³ - a³ + a + 3 = 0

==> a + 3 = 0

==> a = -3

Therefore, the value of 'a' is -3.

VERIFICATION:

p(x) = x³ - ax² + x + 3 = 0

(-3)³ - (-3)(-3)² + (-3) + 3 = 0

-27 - (-3)(9) - 3 + 3 = 0

==> -27 + 27 = 0

==> 0 = 0

Hence, verified!