Question

can someone pls solve this some step by step
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Answer 1

1st equation is $ax^{3} + 3x^2-3$ and the factor is x-4

So f(4)= $4^3a+3(4^2)-3=r1$

          r1= 64a+45

2nd equation is $2x^3-5x+a$ and factor is also x-4

So f(4)=128-20+a=r2

      r2=108+a

From given we know that 2r1=r2

So 2(64a+45)=108+a

        128a+90=108+a

         127a=18

             a=18/127

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