Use the GCF to factor 32x+56

Mathematics

1 answer:

valkas [14]3 years ago

8 0

Answer:

Step-by-step explanation:

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What is 3x² + 3x - 36 and x² - 3x - 28 in factored form?

satela [25.4K]

Answer:

3x² + 3x - 36=

(3x+12)(x-3)

x² - 3x - 28=

(x+4)(x-7)

Step-by-step explanation:

For both equations use the method called the "Cross Method". It is useful for these type of factorisation questions called "Trinomials".

For more info on the cross method.

Here it is:

https://www.mathsteacher.com.au/year10/ch10_factorisation/05_cross_mult_method/cross.htm

Hope you enjoyed :D

3 0

3 years ago

Read 2 more answers

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$