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aleksklad [387]
1 year ago
8

F(x) = 3x^3 - 2x^2 - 11x + 10; (x + 2) I need to factor and find the zeros.

Mathematics
1 answer:
Ratling [72]1 year ago
3 0

Answer:

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

\textsf{Zeros}: \quad x=-2, \quad x=\dfrac{5}{3},\quad x=1

Step-by-step explanation:

<u>Given function</u>:

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 <em>a</em>, compare the coefficients of x³:

\implies a=3

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

\implies 2a+b=-2

\implies 2(3)+b=-2

\implies b=-8

To find <em>c</em>, 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)

<u>Zero Product Property</u>

\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=\dfrac{5}{3}

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

Therefore, the zeros of the function are:

x=-2, \quad x=\dfrac{5}{3},\quad x=1

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Term 2: 200
Term 3: 40

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The rule to find the n^{th} term in a sequence is 
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So, the formula for the sequence in question is
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The sequence is a divergent one. We can always find the value of the next term by dividing the previous term by 5 and if we do that, the value of the next term will get closer to 'zero' but never actually equal to zero.

We can find a partial sum of the sequence using the formula
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