Find the zeros for each function
1. f(x)= x^3 + 12x^2 + 21x + 10

2. f(x) = x^4 + x^3 - 11x^2 - 9x + 18

9514 1404 393

Answer:

Step-by-step explanation:

For higher-degree polynomials I find it extremely useful to use a graphing calculator to find the zeros.

The usual approaches are ...

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1. Rational roots will be factors of 10, so ±1, ±2, ±5, or ±10. The signs are all positive, so there are no positive real roots. The sum of odd-degree coefficients is the same as the sum of even-degree coefficients, so -1 is one of the roots. Removing that using synthetic division gives you ...

   f(x) = (x +1)(x^2 +11 +10)

The quadratic (more or less obviously) factors as (x+1)(x+10), so the complete factorization is ...

  f(x) = (x +1)^2(x +10)

and the zeros are {-10, -1, -1}.

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2. Rational roots will be factors of 18, so will be ±1, ±2, ±3, ±6, ±9, ±18. The sum of coefficients is 0, so x=1 is a root. The rule of signs tells you that there are 2 positive real roots, so there is one more positive real root. Synthetic division tells you the factoring so far is ...

  f(x) = (x -1)(x^3 +2x^2 -9x -18)

We note that the first two coefficients have the same ratio as the last two, so we can factor by grouping.

  x^3 +2x^2 -9x -18 = x^2(x +2) -9(x +2) = (x^2 -9)(x +2)

Factoring the difference of squares gives you the complete factorization, hence all of the roots. (The zeros are the values of x that make the factors zero.)

  f(x) = (x -1)(x +2)(x -3)(x +3)

The zeros are {-3, -2, 1, 3}.

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4 + 2 – (5 – 3)


4 + 2 – (5 + 3)


4 + 2 – (– 5) – (– 3)


4 + 2 + (– 5) + (– 3)

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O 73 rotations
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b.
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c.
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d.
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A
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B
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C
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D
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5
A. 3,467 x
B. 3,467 x
C. 3,467 x
D. 3,467 x
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Find the solutions by using the quadratic formula: *
y=x2 – 5x +11

8w < 32 can u guys please help me

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