Question

If the product of zeros of the polynomial p(x) = ax3 + 19x2 + 38x + 21 is 7 , then the value of a is

a) 3
b)-3
c) 0
d) none of these ​

Answer 1

Given:-

• $p(x) = a {x}^{3} + 19 {x}^{2} + 38x + 21$
• the product of zeros of the polynomial p(x) = ax³ + 19x² + 38x + 21 is 7.

To Find:-

• the value of a.

Solution:-

$\alpha + \beta + \gamma = \frac{ - b}{a }$

$\alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a}$

$\alpha \beta \gamma = \frac{ - d}{a}$

Now, we know that the Product is 7.

So,

$7 = - \frac{19}{a}$

$= > a = \frac{ - 19}{7}$

∴Value of a is -19/7.

Answer 2

Answer => -3

Explanation ,

$\alpha \beta \gamma = \frac{ - d}{a} = 7 \\ = > a = a \: \: \: \: \: \\ b = 12 \: \: \: \: \: \\ c = 38 \: \: \: \: \: \: \: \\ d =21 \\ putting \: the \: values \\ = > \frac{ - 21}{a} = 7 \\ = > - 3$

Hope it helps :)