Question

find the value of a and b so that the p(x) x³-ax²-13x+b has (x-1) and (x+3) as factors plz it's urgent

Answer 1

value of a=3 & value of b=15

Answer 2

x=1, x=(-3)
substitute 'x=1' in p(x),
x^3 - ax^2 - 13x + b =0
(1)^3 - a(1)^2 - 13(1) +b =0
1 - a - 13 +b =0
0= 12 + a - b
0 = a - b + 12 [eq. 1]

(-3)^3 - a(-3)^2 - 13(-3) +b=0
-27 - 9a + 39 +b =0
0= 9a - b - 12 [eq. 2]

a - b + 12=0
9a - b - 12=0
(-) (+) (+)
-8a + 24 =0
a = -24 ÷ -8
a = 3
substitute 'a' in eq 1
a - b +12=0
3 - b +12=0
3+ 12 = b
15 = b
a= 3, b= 15
if it helps plz make my answer brainliest...