Question

algebra ii - semester b / unit 6 - polynomials quiz 2 1. choose the correct roots for each polynomial equation. ( x^3 - 2x + 6 = (x - 1)(x + 3)(x - 2) ) ( circ x=-1,-3,-2 ) ( circ x=1,-3,2 ) ( circ x=1,3,2 ) ( circ x=-1,3,-2 ) 2. choose where the graph of the polynomial functions would cross the ( x )-axis. ( f(x) = x^2 - 4x - 12 ) ( circ x=1,12 ) ( circ x=-1,12 ) ( circ x=-3,4 ) ( circ x=-2,6 ) 3. choose where the graphs of the polynomial functions would cross the ( x )-axis. ( f(x) = 3x^2 - 12x - 36 ) ( circ x=2 ) ( circ x=-6,2 )

Explanation:

Step1: Set polynomial to 0

$x^3 -7x +6=(x-1)(x+3)(x-2)=0$

Step2: Solve for each factor

$x-1=0 \implies x=1$; $x+3=0 \implies x=-3$; $x-2=0 \implies x=2$

Step1: Set polynomial to 0

$f(x)=x^2-4x-12=0$

Step2: Factor the quadratic

$(x-6)(x+2)=0$

Step3: Solve for each factor

$x-6=0 \implies x=6$; $x+2=0 \implies x=-2$

Step1: Set polynomial to 0

$f(x)=3x^2-12x-36=0$

Step2: Simplify the equation

Divide by 3: $x^2-4x-12=0$

Step3: Factor the quadratic

$(x-6)(x+2)=0$

Step4: Solve for each factor

$x-6=0 \implies x=6$; $x+2=0 \implies x=-2$

Answer:

$x=1,-3,2$ (corresponding to the option: $\boldsymbol{x = 1, -3, 2}$)

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