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3 years ago

7

Factor this trinomial completely.

1 answer:

WITCHER [35]3 years ago

8 0

Answer:

Option C. (2x+3)(6x – 5) is the correct answer.

Step-by-step explanation:

Given polynomial(trinomial) is:

$12x^2 + 8x - 15$

The given trinomial is quadratic. Factorization will be as follows:

Multiplying the co-efficient of x^2 and the constant results in 180. We have to make factors of 180 in such a way that their sum of difference is equal to the co-efficient of x.

The required factors are 18 and 10

So,

$=12x^2+18x-10x-15\\=6x(2x+3)-5(2x+3)\\= (6x-5)(2x+3)$

Hence,

Option C. (2x+3)(6x – 5) is the correct answer.

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

(-2) with an exponent of four equals what

charle [14.2K]

The answer is 16 because (-2*-2*-2*-2= 16)

6 0

3 years ago

Write the word name for each number and tell the value of the underlined digit 3,552,308,725 and the number under lined is the s

Alchen [17]

This number is three billion, five hundred fifty-two million, three hundred eight thousand, seven hundred twenty-five.
The second 5 (from the left) is in the ten millions place. It has a value of 50,000,000.

8 0

3 years ago

What the answer is? Please answer fast!

gulaghasi [49]

Answer:

A

Step-by-step explanation:

it includes all the combinations.

4 0

3 years ago

The Earth has a mass of about 6 x 1024 kg. Neptune has a mass of 1 x 1026 kg.

Leno4ka [110]

Neptune is about four times wider than Earth.

8 0

3 years ago