If (x-1) is a factor of Ax^3 + Bx^2 - 36x + 2 | vedclass

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(C) Given that $(x-1)$ is a factor of the polynomial $P(x) = Ax^3 + Bx^2 - 36x + 22$.By the Factor Theorem,$P(1) = 0$.Substituting $x=1$ into the poly

Vedclass · Accepted Answer

$A=2, B=12$

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(C) Given that $(x-1)$ is a factor of the polynomial $P(x) = Ax^3 + Bx^2 - 36x + 22$.By the Factor Theorem,$P(1) = 0$.Substituting $x=1$ into the polynomial:$A(1)^3 + B(1)^2 - 36(1) + 22 = 0$$A + B - 36 + 22 = 0$$A + B = 14$  --- (Equation $1$)Given the second condition: $2^B = 64^A$.Since $64 = 2^6$,we can write:$2^B = (2^6)^A$$2^B = 2^{6A}$Equating the exponents:$B = 6A$ --- (Equation $2$)Substitute Equation $2$ into Equation $1$:$A + 6A = 14$$7A = 14$$A = 2$Now,find $B$ using Equation $2$:$B = 6(2) = 12$Therefore,$A=2$ and $B=12$.

If $(x-1)$ is a factor of $Ax^3 + Bx^2 - 36x + 22$ and $2^B = 64^A$,find $A$ and $B$.

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