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valentina_108 [34]
2 years ago
6

Determine the number and type of roots for the equation using one of the given roots. Then find each root. (inclusive of imagina

ry roots.) 1. x^3-7x+6=0;1 2. x^3-3x^2+25x+29=0;-1 3.x^3-4x^2-3x+18=0;3 Find all the zeros of the function 4. f(x)=x^2+4x-12 5.f(x)=x^3-3x^2+x+5 6. f(x)=x^3-4x^2-7x+10 Write the simplest polynomial function with integral coefficients that has the given zeros. 7. -5,-1,3,7 8. 4,2+3i (thank you so much in advance, it would mean a lot, and it is urgently needed, hence the high reward, thank you! Have a great day)
Mathematics
1 answer:
dmitriy555 [2]2 years ago
5 0

Step-by-step explanation:

<em>"Determine the number and type of roots for the equation using one of the given roots. Then find each root. (inclusive of imaginary roots.)"</em>

Given one of the roots, we can use either long division or grouping to factor each cubic equation into a binomial and a quadratic.  I'll use grouping.

Then, we can either factor or use the quadratic equation to find the remaining two roots.

1. x³ − 7x + 6 = 0; 1

x³ − x − 6x + 6 = 0

x (x² − 1) − 6 (x − 1) = 0

x (x + 1) (x − 1) − 6 (x − 1) = 0

(x² + x − 6) (x − 1) = 0

(x + 3) (x − 2) (x − 1) = 0

The remaining two roots are both real: -3 and +2.

2. x³ − 3x² + 25x + 29 = 0; -1

x³ − 3x² + 25x + 29 = 0

x³ − 3x² − 4x + 29x + 29 = 0

x (x² − 3x − 4) + 29 (x + 1) = 0

x (x − 4) (x + 1) + 29 (x + 1) = 0

(x² − 4x + 29) (x + 1) = 0

x = [ 4 ± √(16 − 4(1)(29)) ] / 2

x = (4 ± 10i) / 2

x = 2 ± 5i

The remaining two roots are both imaginary: 2 − 5i and 2 + 5i.

3. x³ − 4x² − 3x + 18 = 0; 3

x³ − 4x² − 3x + 18 = 0

x³ − 4x² + 3x − 6x + 18 = 0

x (x² − 4x + 3) − 6 (x − 3) = 0

x (x − 1)(x − 3) − 6 (x − 3) = 0

(x² − x − 6) (x − 3) = 0

(x − 3) (x + 2) (x − 3) = 0

The remaining two roots are both real: -2 and +3.

<em>"Find all the zeros of the function"</em>

For quadratics, we can factor using either AC method or quadratic formula.  For cubics, we can use the rational root test to check for possible rational roots.

4. f(x) = x² + 4x − 12

0 = (x + 6) (x − 2)

x = -6 or +2

5. f(x) = x³ − 3x² + x + 5

Possible rational roots: ±1/1, ±5/1

f(-1) = 0

-1 is a root, so use grouping to factor.

f(x) = x³ − 3x² − 4x + 5x + 5

f(x) = x (x² − 3x − 4) + 5 (x + 1)

f(x) = x (x − 4) (x + 1) + 5 (x + 1)

f(x) = (x² − 4x + 5) (x + 1)

x = [ 4 ± √(16 − 4(1)(5)) ] / 2

x = (4 ± 2i) / 2

x = 2 ± i

The three roots are x = -1, x = 2 − i, x = 2 + i.

6. f(x) = x³ − 4x² − 7x + 10

Possible rational roots: ±1/1, ±2/1, ±5/1, ±10/1

f(-2) = 0, f(1) = 0, f(5) = 0

The three roots are x = -2, x = 1, and x = 5.

<em>"Write the simplest polynomial function with integral coefficients that has the given zeros."</em>

A polynomial with roots a, b, c, is f(x) = (x − a) (x − b) (x − c).  Remember that imaginary roots come in conjugate pairs.

7. -5, -1, 3, 7

f(x) = (x + 5) (x + 1) (x − 3) (x − 7)

f(x) = (x² + 6x + 5) (x² − 10x + 21)

f(x) = x² (x² − 10x + 21) + 6x (x² − 10x + 21) + 5 (x² − 10x + 21)

f(x) = x⁴ − 10x³ + 21x² + 6x³ − 60x² + 126x + 5x² − 50x + 105

f(x) = x⁴ − 4x³ − 34x² + 76x − 50x + 105

8. 4, 2+3i

If 2 + 3i is a root, then 2 − 3i is also a root.

f(x) = (x − 4) (x − (2+3i)) (x − (2−3i))

f(x) = (x − 4) (x² − (2+3i) x − (2−3i) x + (2+3i)(2−3i))

f(x) = (x − 4) (x² − (2+3i+2−3i) x + (4+9))

f(x) = (x − 4) (x² − 4x + 13)

f(x) = x (x² − 4x + 13) − 4 (x² − 4x + 13)

f(x) = x³ − 4x² + 13x − 4x² + 16x − 52

f(x) = x³ − 8x² + 29x − 52

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