Mathematics, 26.09.2019 10:10, ctdavis

Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and coefficients 6x^2-3-7x

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answered: Jessicawilliams3569

see explanation

Step-by-step explanation:

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

Express in standard form, that is 6x² - 7x - 3

product = 6 × - 3 = - 18, sum = - 7

The factors are - 9 and + 2

Use these factors to split the middle term

6x² - 9x + 2x - 3 ( factor the first/second and third/fourth terms )

= 3x(2x - 3) + 1(2x - 3) ← factor out (2x - 3)

= (2x - 3)(3x + 1)

To obtain zeros equate to zero

(2x - 3)(3x + 1) = 0

Equate each factor to zero and solve for x

2x - 3 = 0 ⇒ x = $\frac{3}{2}$

3x + 1 = 0 ⇒ x = - $\frac{1}{3}$

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The relationship between the zeros and the coefficients is

sum of zeros = - $\frac{b}{a}$ and

product of zeros = $\frac{c}{a}$

- $\frac{b}{a}$ = - $\frac{7}{6}$ and

$\frac{3}{2}$ - $\frac{1}{3}$ = $\frac{7}{6}$

$\frac{3}{2}$ × - $\frac{1}{3}$ = - $\frac{1}{2}$ and

$\frac{c}{a}$ = $\frac{-3}{6}$ = - $\frac{1}{2}$

Verifying both relationships

answered: Guest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2: