Question

Write a polynomial function, p(x) with degree 3 that has p(7)=0

Answer 1

Answer:

$p (x) = x^{3} - 21x^{2}+ 147x - 343$

is the required polynomial with degree 3 and p ( 7 ) = 0

Step-by-step explanation:

Given:

p ( 7 ) = 0

To Find:

p ( x ) = ?

Solution:

Given p ( 7 ) = 0 that means substituting 7 in the polynomial function will get the value of the polynomial as 0.

Therefore zero's of the polynomial is seven i.e 7

Degree : Highest raise to power in the polynomial is the degree of the polynomial

We have the identity,

$(a -b)^{3} = a^{3}-3a^{2}b +3ab^{2} - b^{3}$

Take a = x

        b = 7

Substitute in the identity we get

$(x -7)^{3} = x^{3}-3x^{2}(7) +3x(7)^{2} - 7^{3}\\(x -7)^{3} = x^{3}-21x^{2} +147x - 343$

Which is the required Polynomial function in degree 3 and if we substitute 7 in the polynomial function will get the value of the polynomial function zero.

p ( 7 ) = 7³ - 21×7² + 147×7 - 7³

p ( 7 ) = 0

$p (x) = x^{3} - 21x^{2}+ 147x - 343$