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4 years ago

Factor the Higher degree polynomial 5y^4 + 11y^2 + 2

Mathematics

2 answers:

kotykmax [81]4 years ago

8 0

$\bf 5y^4+11y^2+2\implies 5(y^2)^2+11y^2+2\implies (5y^2+1)(y^2+2)$

RideAnS [48]4 years ago

5 0

For this case we must factor the following polynomial:

$5y ^ 4 + 11y ^ 2 + 2$

We rewrite $y ^ 4$ as $(y^ 2) ^ 2$ :

$5 (y ^ 2) ^ 2 + 11y ^ 2 + 2$

We make a change of variable:

$u = y ^ 2$

We replace:

$5u ^ + 11u + 2$

we rewrite the middle term as a sum of two terms whose product of 5 * 2 = 10 and the sum of 11.

So:

$5u ^ 2 + (1 + 10) u + 2$

We apply distributive property:

$5u ^ 2 + u + 10u + 2$

We factor the highest common denominator of each group.

$(5u ^ 2 + u) + (10u + 2)\\u (5u + 1) +2 (5u + 1)$

We factor again:

$(u + 2) (5u + 1)$

Returning the change:

$(y ^ 2 + 2) (5y ^ 2 + 1)$

ANswer:

$(y ^ 2 + 2) (5y ^ 2 + 1)$

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4 years ago

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1. Determinar la ecuación canónica de la parábola con vértice en (-2,4) y foco en (1,4) 2. Determinar el foco y el vértice de la

goldenfox [79]

Answer:

1. La ecuación de la parábola en forma canónica es x = 1/12 × (y - 4) ² - 2

2. Vértice = (-1, 3), enfoque = (-5/2, 3)

3. y = 12x no es una parábola

4. y = -8x, no es una parábola

Step-by-step explanation:

1. La ecuación estándar de una parábola es y = a · x² + b · x + c

El vértice V es (h, k)

El foco (h + p, k)

Por lo tanto, tenemos en comparación k = 4, h = -2

h + p = 1

p = 1 - h = 1 - (-2) = 3

Lo que da la ecuación como (y - k) ² = 4 · p · (x - h)

Al ingresar los valores de k, h y p, tenemos

(y - 4) ² = 4 × 3 × (x - (-2)) = 12 × (x + 2)

12 · x + 24 = (y - 4) ²

x = 1/12 × (y - 4) ² - 2

La ecuación de la parábola en forma canónica es x = 1/12 × (y - 4) ² - 2

2. Determinar el foco y el vértice de la parábola (y - 3) ² = -6 · (x + 1)

Reescribimos la ecuación en forma de vértice de la siguiente manera;

-6 · x -6 = (y - 3) ²

x = -1 / 6 × (y - 3) ² - 1

La ecuación de una parábola en forma de vértice es x = a · (y - k) ² + h

Con el vértice = (h, k)

Comparando, tenemos, h = -1 yk = 3, el vértice = (-1, 3)

También la ecuación de la parábola en forma cónica es (y - k) ² = 4 · p · (x - h)

Comparando con (y - 3) ² = -6 · (x + 1), tenemos 4p = -6, p = -3/2

El foco está en (h + p, k) que es (-1 + -3/2, 3) = (-5/2, 3)

Vértice = (-1, 3), Enfoque = (-5/2, 3)

3. Para la parábola, y = 12 · x, tenemos;

En comparación con la forma de la ecuación, y = a · x² + b · x + c

b = 12, a = 0, c = 0

Dado que el vértice = (h, k), tenemos;

h = -b / (2 × 0), h = ∞

k = a · h² + b · h + c = ∞

No hay vértice

Foco x valor = Vértice x valor = ∞

No hay foco

Directrix = (k - 1) / (4 · a) = (k - 1) / (4 × 0) = ∞, sin directriz

y = 12x no es una parábola

4. Para y = -8x, tampoco es una parábola como se muestra arriba.

4 0

3 years ago

In this diagram, ABAC – AEDF. If the<br> area of ABAC = 6 in?, what is the<br> area of AEDF?

shepuryov [24]

Answer:

2.7 in²

Step-by-step explanation:

similar triangles have the same angles, and all side lengths (or other distances) of one triangle have the same scaling factor to the side lengths of the other triangle.

so, we know the relation between the 2 baselines is 2/3, as this is the factor to turn the baseline of the large triangle into the baseline of the smaller triangle.

in other words

EF = BC × 2/3

2 = 3 × 2/3

correct

how do we calculate the area of a triangle ?

Area = baseline × height / 2

from BAC we know

Area = 6

baseline = 3

height = ?

6 = 3 × height / 2

12 = 3 × height

height = 4

aha !

now, EDF has a height too that we need to calculate is Area. and this height has the same scaling factor compared to the larger triangle as the side lengths : 2/3

so, for EDF we know

Area = ?

baseline = 2

height = 4 × 2/3 = 8/3

therefore, the area is

Area = (2 × 8/3) / 2 = (16/3) / 2 = 8/3 = 2.66666... ≈ 2.7

the shirt answer would be :

we know from the 2 baselines that the scaling factor for each distance is 2/3.

for the area we need to multiply 2 distances, so that means we have to multiply both by 2/3. and so on the formula for the area we have to use 2/3 × 2/3.

2/3 × 2/3 = 4/9

Area small = Area large × 4/9 = 6 × 4/9 = 24/9 = 8/3 ≈ 2.7

3 0

3 years ago

Write equations for the horizontal and vertical lines passing through the point (6,9)

Maslowich

The equations for the horizontal and vertical lines passing through the point (6,9) is y = 9 and x = 6 respectively

<u>Solution:</u>

Given, point is (6, 9)

We have to find the equations for horizontal and vertical lines passing through above given point.

Now, let us find horizontal line,

We know that, horizontal line is parallel to x – axis, so slope of our required line is 0.

The point slope form is given as $y - y_1 = m(x - x_1)$

Then, line equation in point slope form ⇒ y – 9 = 0(x – 6)

⇒ y – 9 = 0

⇒ y = 9

Now, let us find vertical line,

We know that, vertical line is parallel to y – axis, so slope of our required line is undefined $(\frac{1}{0})$

Then, line equation in point slope form ⇒ $y-9=\frac{1}{0}(x-6)$

⇒ x – 6 = 0(y – 9)

⇒ x – 6 = 0

⇒ x = 6

Hence, the horizontal line equation is y = 9 and vertical line equation is x = 6.

7 0

3 years ago