Question

find a cubic polynomial whose zeros are 4 - 3 and -1

Answer 1

Answer:

x³-13x-12

Step-by-step explanation:

Any cubic polynomial with leading coefficient 'a' and zeros 'm', 'n' and 'p'

will be of the form

a(x-m)(x-n)(x-p).

Now given zeros of the cubic polynomial to be

4,-3,-1

Hence, cubic polynomial will be in the form

a(x-4)(x+3)(x+1)

=a(x³-13x-12)

Now, suppose if we assume the leading coefficient to be 1 then the cubic

polynomial will be of the form

=x³-13x-12

Answer 2

Solution:

Another method to find the cubic polynomial when all zeros are given

As we know that standard equation of cubic polynomial

$a {x}^{3} + b {x}^{2} + cx + d \\ \\ {x}^{3} - ( \frac{ - b}{a} ) {x}^{2} + ( \frac{c}{a} )x - ( - \frac{d}{a} )$
coefficient and zeros have relation as

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \\ \\ \alpha \beta \gamma = \frac{ - d}{a} \\ \\ \alpha = 4 \\ \\ \beta = - 3 \\ \\ \gamma = - 1$
$4 - 3 - 1 = 0 = \frac{ - b}{a} \\ \\ - 12 + 3 - 4 = - 13 = \frac{c}{a} \\ \\ 12 = \frac{ - d}{a}$
So cubic polynomial is

${x}^{3} - (0 ) {x}^{2} + ( - 13)x - ( 12) \\ \\ = > {x}^{3} - 13x - 12$
is the final answer.