The function f is defined as f(x) = x(x² - 4) - 3x(x - 2).
Part A
f(x) = x(x² - 4) - 3x(x - 2) and f(x) = x(x - 2)(x - a) are equivalent. Find the value of a. Therefore, x(x² - 4) - 3x(x - 2) = x(x - 2)(x - a) {set the functions equal to each other} x(x + 2)(x - 2) - 3x(x - 2) = x(x - 2)(x - a) {factored the x² - 4 into (x + 2)(x - 2)} x + 2 - 3 = x - a {divided each side by x(x - 2)} x - 1 = x - a {combined like terms} -a = -1 {subtracted x from each side} a = 1 {divided each side by -1} Part B Which values are zeros of the function, f? f(x) = x(x - 2)(x - 1) {replaced a with 1 in the equivalent form, because it is already factored} x(x - 2)(x - 1) = 0 {substituted 0, for f(x), to find the zeros} x = 0 or x - 2 = 0 or x - 1 = 0 {set each factor equal to 0} x = 0 or x = 2 or x = 1 {solved each equation for x} D.) 0, E.) 1, and F.) 2 are the zeros Ask Algebra House Comments are closed.
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