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

Step-by-step explanation:

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

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.

"Find all the zeros of the function"

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.

"Write the simplest polynomial function with integral coefficients that has the given zeros."

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

the answer is 16ft

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!