Question

If -1 is a zero of p(x)=ax^3+x^2+x+4, then find the value of a​

Answer 1

Given equation

$ax^{3} +x^{2} +x+4=0$

-1 is the zero of the polynomial

Substituting x = -1, we get,

$\implies a(-1)^{3}+(-1)^{2} +(-1)+4=0$

( A negative number raised to an odd number is negative

A negative number raised to an even number is positive)

$\implies -a+1-1+4=0$

$\implies -a+0+4=0$

Transposing a number to the other side of the equal sign, changes the sign.

$\implies -a=-4$

$\implies a = 4$

$\boxed{\boxed{\bold{Therefore, \ the \ value \ of \ a \ is \ 4}}}}$

                                                           

Answer 2

GIVEN :-

(-1) is a zero of the polynomial p(x) = ax³ + x² + x + 4.

TO FIND :-

The value of 'a'.

HERE'S WHAT WE NEED TO KNOW :-

If a zero of a polynomial is given, then if the value of the variable is replaced by the zero, then the result will be 0.

For example, is (1) is a zero of p(x), then p(1) = 0.

SOLUTION ;-

Zero of the polynomial = (-1)

Now,

p(-1) = a(-1)³ + (-1)² + (-1) + 4 = 0

⇒ -a + 1 - 1 + 4 = 0

⇒4 - a = 0

⇒a = 4

Therefore, the value of 'a' is 4.