Question

Find the values of a and b so that the polynomial x∆3 - ax∆2 - 13x +b has x-1 and x+3 as factors

Answer 1

Step by step explaination :-

x-1=0 , x+3=0

x=1 , x= -3

p(x)=x3-ax2-13x+b

p(1)=(1)3-a(1)2-13×1+b

= 1-a-13+b

= a-12+b

= a=12-b

p(-3)=(-3)3-a(-3)2-13(-3)+b

= -27-9a+39+b

= 12-9a+b

= b=9a-12

b=9(12-b)-12

b=108-9b-12

b=96-9b

b+9b=96

10b=96

b=96/10

b=9.6

a=12-b

a=12-9.6

a= 2.4

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