Question

The polynomial ax³ + 3x²-26. and 2x³ - 5 + a; when divided by (x-4) leave the remainders R1, and R2 respectively. Find the value of a, if R1+R2=0​

Answer 1

Answer:

we have to substitute the values mate

Step-by-step explanation:

$(x-4)=0\\x=4\\Now ,f(x)=ax^3+3x^2-26=0\\f(4)=a(4)^3+3(4)^2-26=0\\64a+19-26=0\\64a-7=0(R1)\\\\Now ,g(x)=2x^3-5+a=0\\g(4)=2(4)^3-5+a=0\\66-5+a=0\\61+a=0(R2)\\\\THEREFORE,\\R1+R2=0\\(64a-7)+(61+a)=0\\65a+54=0\\65a=-54\\a=-54/65$