Please can you fully factorise these equasions?

2 answers:

5 0

$1)\\ 36x^3-6x=6x(6x^2-1)=6x[( \sqrt{6} \cdot x)^2-1^2]=6x(x \sqrt{6} -1)(x \sqrt{6} +1)\\ \\2)\\12x^3+15x^2=3x^2(4x+3)\\ \\3)\\8x-10x^2=2x(4-5x)$

Murljashka [212]3 years ago

3 0

$36x^3-6x=6x(6x^2-1)=6x[(x\sqrt6)^2-1^2]=6x(x\sqrt6-1)(x\sqrt6+1)\\\\12x^3+15x^2=3x^2(4x+5)\\\\8x-10x^2=2x(4-5x)$

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2/5x = 16 what’s x????

VladimirAG [237]

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4 0

3 years ago

Please help ASAP!

Lana71 [14]

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8 0

4 years ago

I’m working on 3 and 4 only maybe if I can have some explanation so I can understand and do more of these next week.

guapka [62]

4.b.

Answer: See below.

Step-by-step explanation:

<h2>For the equation f(x) = 2x</h2>

3.a. f(6) means use x = 6 in the equation f(x) = 2x

so f(6) would be f(6)= 2(6)

f(6) = 12

3.b. f(-11) = 2(-11)

f(-11) = -22

3.c. f(2.75) = 2(2.75)

f(2.75) = 5.5

3.d. This is turned around. We are told f(x)=20, so what would x need to be for f(x) to be 20? Since f(x) = 2x, we can say 20 = 2x. Therefore x = 10

f(10) = 20

The rest of (3) are solved in the same fasion.h

<h2>For the equation f(x)= 5x+50</h2>

4.a. f(7) = 5(7)+50

f(7) = 85

4.b. f(-12)

f(-12) = 5*(-12)+50

f(-12) = -60

Continue in the same fashion for these types of problems.

7 0

2 years ago

What is the answer to -2a-9=6a+15