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F(x) = 3x^3 - 2x^2 - 11x + 10; (x + 2)

I need to factor and find the zeros.

1 Answer

6 votes

Answer:


\textsf{Factored function}: \quad f(x)=(x+2)(3x-5)(x-1)


\textsf{Zeros}: \quad x=-2, \quad x=(5)/(3),\quad x=1

Explanation:

Given function:


f(x)=3x^3-2x^2-11x+10

If (x + 2) is a factor then:


\implies f(x)=(x+2)(ax^2+bx+c)

Expand:


\implies f(x)=ax^3+bx^2+cx+2ax^2+2bx+2c


\implies f(x)=ax^3+2ax^2+bx^2+2bx+cx+2c


\implies f(x)=ax^3+(2a+b)x^2+(2b+c)x+2c

To find a, compare the coefficients of x³:


\implies a=3

To find b, substitute the found value of a into the coefficient for x² and compare:


\implies 2a+b=-2


\implies 2(3)+b=-2


\implies b=-8

To find c, compare the constants:


\implies 2c=10


\implies c=5

Therefore:


\implies f(x)=(x+2)(3x^2-8x+5)

Now factor (3x²-8x+5):


\implies 3x^2-3x-5x+5


\implies 3x(x-1)-5(x-1)


\implies (3x-5)(x-1)

Therefore the factored function is:


\implies f(x)=(x+2)(3x-5)(x-1)

Zero Product Property


\boxed{\text{If \; $a \cdot b = 0$ \; then either \; $a = 0$ \;or \;$b = 0$ (or both)}.}

To find the zeros, set each factor equal to zero and solve for x:


\implies x+2=0 \implies x=-2


\implies 3x-5=0 \implies x=(5)/(3)


\implies x-1=0 \implies x=1

Therefore, the zeros of the function are:


x=-2, \quad x=(5)/(3),\quad x=1

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