Mathematics, 31.10.2019 13:31 naomicervero

Give the solutions from this questions. 50 pointss

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Answer from: masontdavis

7. P(x) = $\bold{\dfrac{3}{4}}$ (x³ 1 x² - 10x + 6)

Step-by-step explanation:

The factored form of a polynomial is: a(x - p)(x-q)...(x-z) . where p, q, ..., z are the roots.

Convert the roots into factors, include conjugates of any complex roots, and multiply all of the factors. Then find the a-value.

roots: x = 3 x = 2 + √2 x = 2 - √2 (conjugate)

factors: x - 3 = 0 x - 2 - √2 = 0 x - 2 + √2 = 0

(x - 3)(x - 2 - √2)(x - 2 + √2)

multiply the complex roots:

(x + 3)(x² - 4x + 2)

distribute (x + 3) to find the basic polynomial

x³ + x² - 10x + 6

Next, find the a-value:

P(x) = a(x³ + x² - 10x + 6)

P(1) = a(1 - 1 - 10 + 6) = -3

-4a = -3

$a=\dfrac{3}{4}$

$\boxed{P(x)=\dfrac{3}{4}(x^3-x^2-10x+6)}$

***********************************************************************************

8. P(x) = $\bold{\dfrac{1}{2}}$ (x⁴ + 8)

Step-by-step explanation:

The factored form of a polynomial is: a(x - p)(x-q)...(x-z) . where p, q, ..., z are the roots.

Convert the roots into factors, include conjugates of any complex roots, and multiply all of the factors. Then find the a-value.

roots: x = -2 x = 1 + i x = 1 - i (conjugate)

factors: x + 2 = 0 x - 1 - i = 0 x - 1 + i = 0

The polynomial has a degree of 4, so x = -2 must be a double root.

(x + 2)²(x - 1 - i)(x - 1 + i)

multiply the double root & complex roots:

(x² + 4x + 4)(x² - 2x + 2)

distribute (x² + 4x + 4) to find the basic polynomial

x⁴ + 8

Next, find the a-value:

P(x) = a(x⁴ + 8)

P(0) = a(0 + 8) = 4

8a = 4

$a=\dfrac{1}{2}$

$\boxed{P(x)=\dfrac{1}{2}(x^4+8)}$

***********************************************************************************

9. P(x) = $\bold{-\dfrac{1}{2}}$ (x⁴ - x³ - 6x²)

Step-by-step explanation:

The factored form of a polynomial is: a(x - p)(x-q)...(x-z) . where p, q, ..., z are the roots. The roots are where the graph crosses the x-axis.

roots: x = -2 x = 0 x = 3

factors: x + 2 = 0 x = 0 x - 3 = 0

The graph "touches - not crosses" the x-axis at x = 0 so x = 0 is a double root.

(x + 2)(x)²(x - 3)

multiply the factors (x + 2)(x - 3):

x²(x² - x - 6)

distribute (x)² to find the basic polynomial

x⁴ - x³ - 6x²

Next, find the a-value:

P(x) = a(x⁴ - x³ - 6x²)

P(2) = a[(2)⁴ -(2)³ - 6(2)²] = 8

a(16 - 8 - 24) = 8

-16a = 8

$a=-\dfrac{1}{2}$

$\boxed{P(x)=-\dfrac{1}{2}(x^4-x^3-6x^2)}$

***********************************************************************************

10. P(x) = -(x³ - 2x² + x - 2)

Step-by-step explanation:

The graph displayed is an x³ graph that has been shifted to the left 2 units and reflected over the x-axis. Only one root is shown, which means the other two roots are complex numbers. Since P(i) = 0, then another root is: x = i

roots: x = -2 x = i x = - i (conjugate)

factors: x + 2 = 0 x - i = 0 x + i = 0

(x + 2)(x - i)(x + i)

multiply the complex roots:

(x + 2)(x² + 1)

distribute (x + 2) to find the basic polynomial

x³ - 2x² + x - 2

Next, reflect over the x-axis:

P(x) = - (x³ - 2x² + x - 2)

Answer from: Quest

rick would give her the best deal as he will charge $8.88 per hour

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

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