Question

find the value of a and b so that x3 - ax2 - 13x + b has x-1 and x+3 as factors

Answer 1

polynomial , x³ -ax² -13x + b has two factors ( x -1) and (x +3)

it means , x = 1 and -3 are the roots (zeros ) of x³ -ax² -13x + b .

so,

put x = 1 in polynomial ,.

(1)³ -a(1)² -13(1) + b = 0

1 - a -13 + b = 0

-a + b = 12 ---------(1)

put x = -3

(-3)³ -a(-3)² -13(-3) + b = 0

-27 -9a +39 + b = 0

-9a + b = -12 --------(2)

subtract equation (1) and (2)

-a + b + 9a - b = 12 + 12

8a = 24

a = 3

put a = 3 in equation (1)

b = 15

Answer 2

x³-ax²-13x+b has two factors.
x-1 and x+3
x-1=0
x=1
and
x+3=0
x= -3

Put x=1,
(1)³-a(1)²-13(1)+b=0
1-a-13+b=0
-a+b-12=0
-a+b=12----------------(1)

Put x=-3,
(-3)³-a(-3)²-13(-3)+b=0
-27-a(9)+36+b=0
-27-9a+36+b=0
-9a+b+12=0
-9a+b=-12----------------(2)

(1)-(2)
-a+b-(-9a+b)=12-(-12)
-a+b+9a-b=24
8a=24
a=24/8=3

-a+b=12
-3+b=12
b=12+3
b=15

So, a=3,b=15