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2 years ago

Factor the expression using the GCF (Greatest Common Factor) 12x + 36

Mathematics

1 answer:

navik [9.2K]2 years ago

5 0

Answer:

12 ( x+3)

Step-by-step explanation:

We can factor a 12 out of each term

12 *x + 12*3

12 ( x+3)

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Find the following algebra, I am in class and plz give explanation

Lelechka [254]

1) Solving $4a+4b-ac-bc$ we get $(a+b)(4-c)$

2) Solving $16a^4-8b$ we get $8(2a^4-b)$

3) Solving $x^2-8x+15$ we get $(x-3)(x-5)$

Step-by-step explanation:

We need to solve the following expressions:

1) $4a+4b-ac-bc$

We will solve this expression by factoring common terms:

$4a+4b-ac-bc$

Factoring 4 from 4a+4b and -c from -ac-bc

$4(a+b)-c(a+b)$

Taking common term

$(a+b)(4-c)$

2) $16a^4-8b$

We will solve this expression by factoring common terms:

$16a^4-8b$

Taking 8 as common term:

$8(2a^4-b)$

3) $x^2-8x+15$

We can solve the expression by breaking the middle term into 2

$x^2-3x-5x+15\\Finding\,\,common:\\=x(x-3)-5(x-3)\\=(x-3)(x-5)$

Keywords: Finding factors

Learn more about Finding factors at:

#learnwithBrainly

6 0

2 years ago

Integration of sin^2 2x cos^3 2x dx

lilavasa [31]

$\int\left[\sin^2(2x)\cos^3(2x)\right]dx=\int\left[\sin^2(2x)\cos^2(2x)\cos(2x)\right]dx\\\\\text{Use the trigonometric identity}\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\=\int\left\{\sin^2(2x)[1-\sin^2(2x)]\cos(2x)\right\}dx\\\\\text{substitute}\ \sin2x=t\ \to\ 2\cos2x\ dx=dt\ \to\ \cos2x\ dx=\dfrac{1}{2}\ dt\\\\=\dfrac{1}{2}\int\left[t^2(1-t^2)\right]dt=\dfrac{1}{2}\int(t^2-t^4)dt=\dfrac{1}{2}\left(\dfrac{1}{3}t^3-\dfrac{1}{5}t^5\right)+C$

$=\dfrac{1}{2}\left(\dfrac{1}{3}\sin^3(2x)-\dfrac{1}{5}\sin^5(2x)\right)+C$

4 0

3 years ago

Find value of x 4(x+6)° (5x+11)°

Evgesh-ka [11]

Value of x = 13°

Hope it helps you...

5 0

2 years ago

Please solve this<br><img src="https://tex.z-dn.net/?f=%20%7Cx%20%5E%7B2%7D%20%2B%203x%20-%201%20%7C%20%20%3C%203" id="TexFormul

klasskru [66]

$\bf |x^2+3x-1|\ \textless \ 3\implies 
\begin{cases}
+(x^2+3x-1)\ \textless \ 3\\\\
-(x^2+3x-1)\ \textless \ 3
\end{cases}\\\\
-------------------------------\\\\
x^2+3x-1\ \textless \ 3\implies x^2+3x-4\ \textless \ 0\implies (x+4)(x-1)\ \textless \ 0
\\\\\\
x\ \textless \ 
\begin{cases}
-4\\
1
\end{cases}\\\\
-------------------------------\\\\
-(x^2+3x-1)\ \textless \ 3\implies x^2+3x-1\boxed{\ \textgreater \ }-3
\\\\\\
x^2+3+2\ \textgreater \ 0\implies (x+2)(x+1)\ \textgreater \ 0\implies x\ \textgreater \ 
\begin{cases}
-2\\
-1
\end{cases}$