Question

find the value of p and q so that 1 and-3 are the zeros of the polynomial
$\\ x {}^{3} - px {}^{2} - 13x + q$
$\\ \\ who \: will \: answer \: it \: coretly \: and \: ans \: it \: first \: wll \: be \: marked \: as \: brainlist$
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Answer 1

hope this helps you.. :)

Answer 2

Answer:

p=3,q=15

"You are the first who will receive the correctly answered solution"

Step-by-step explanation:

f(x)=x³-px²-13x+q

1 is the root of f(x)

so f(1)=0

or 1³-p*1²-13*1+q=0

1-p-13+q=0

-p+q=12...........................(1)

Also -3 is root of f(x)

so f(-3)=0

or (-3)³-p(-3)²-13(-3)+q=0

-27-9p+39+q=0

-9p+q=-12-------------------------(2)

subtracting (2) from(1)

-p+9p=12-(-12)

8p=24

p=24/8=3

putting p=3 in (1) we get

-3+q=12

q=12+3=15

hence p=3,q=15