When 4x^3 - ax^2 + bx - 4 is divided by x - 2 | vedclass

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(A) Let $f(x) = 4x^3 - ax^2 + bx - 4$.According to the Remainder Theorem,when $f(x)$ is divided by $(x - c)$,the remainder is $f(c)$.For divisor $(x -

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$a = 3, b = 2$

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(A) Let $f(x) = 4x^3 - ax^2 + bx - 4$.According to the Remainder Theorem,when $f(x)$ is divided by $(x - c)$,the remainder is $f(c)$.For divisor $(x - 2)$,the remainder is $f(2) = 20$:$4(2)^3 - a(2)^2 + b(2) - 4 = 20$$4(8) - 4a + 2b - 4 = 20$$32 - 4a + 2b - 4 = 20$$28 - 4a + 2b = 20$$2b - 4a = -8$Dividing by $2$,we get $b - 2a = -4$,or $2a - b = 4$ ... $(1)$For divisor $(x + 1)$,the remainder is $f(-1) = -13$:$4(-1)^3 - a(-1)^2 + b(-1) - 4 = -13$$4(-1) - a(1) - b - 4 = -13$$-4 - a - b - 4 = -13$$-a - b - 8 = -13$$-a - b = -5$,or $a + b = 5$ ... $(2)$Adding equations $(1)$ and $(2)$:$(2a - b) + (a + b) = 4 + 5$$3a = 9 \Rightarrow a = 3$Substituting $a = 3$ into equation $(2)$:$3 + b = 5 \Rightarrow b = 2$Thus,the values are $a = 3$ and $b = 2$.

When $4x^3 - ax^2 + bx - 4$ is divided by $x - 2$ and $x + 1$,the respective remainders are $20$ and $-13$. Find the values of $a$ and $b$.

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