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3 years ago

6

Is the binomial a factor of the polynomial nominal function?

1 answer:

Sedbober [7]3 years ago

7 0

The factors for given equation f(x)= $x^{3} +3x^{2} -25x-75$ .

(x-1) - No
(x-3) - No
(x+3) - Yes
(x-5) - Yes
(x+5) - Yes

<u>Step-by-step explanation:</u>

The given equation is f(x)= $x^{3} +3x^{2} -25x-75$ .

Add 0 at the end of the equation.

$x^{3} +3x^{2} -25x-75$ = 0.

Let us group the given equation,

$(x^{3} +3x^{2}) -(25x-75)$ =0.

⇒ Group 1: $x^{3} +3x^{2}$ .

Group 2: $- 25x +75$ .

Pull out factor from each group,

⇒ Group 1: $(x+3)(x^{2})$ .

Group 2: (x+3) (-25).

Join the two group since both (x+3) is common in both groups.

$(x+3)(x^{2} -25)$ =0.

One of the factor is (x+3).

Other factors are solved by the formula, $a^{2}-b^{2} = (a+b) (a-b)$ .

$(x+3)(x^{2} -25)$ = (x+5) (x-5) .

The other factors are (x+5) and (x-5).

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Step-by-step explanation:

Lower level = 4x - 300

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Combine the two equations to find the number of seats in the upper level

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3 years ago

The lemonade cooler at the class picnic holds 12.5L. One liter is approximately equal to 0.26 gallons. How many gallons does the

lora16 [44]

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4 years ago

The book rack sold 24 paperbacks on Saturday, 17 of them mysteries. What percent of the books were mysteries?

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Answer:

About 71% of the books were mystery books

Step-by-step explanation:

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3 years ago

alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

$y = \frac{1}{5}x + \frac{12}{5}$

$D = (\frac{43}{26},\frac{71}{26})$

Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

$m = -5$

The slope of the line that passes through B is calculated as:

$m_2 = -\frac{1}{m}$ --- because it is perpendicular to AC.

So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

$5y = x - 3 + 15$

$5y = x +12$

Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

$y=-5x + 10 + 1$

$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

Multiply through by 5

$-25x-x = 12 - 55$

Collect like terms

$-26x = -43$

Solve for x

$x = \frac{-43}{-26}$

$x = \frac{43}{26}$

Substitute $x = \frac{43}{26}$ in $y=-5x + 11$

$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

3 0

3 years ago