Mathematics common factoring?

Question

common factoring negative out

say you have -x^2-x+6>0 you have to common factor - rite but since you divide everything in left side by -1 you have to divide everything on right side by negative one as well? so -(x^2+x-6)<0 and -(x-2)(x+3) <0 rite

second question -3x^2+6x+12=0

so would it be -3(x^2-2x-4)=0

or if it was -3x^2+6x+12=4

it would be -3(x^2-2x-4)=4/(-3)?

Answer 1

i&#039;m not sure about the first one, but for the second one, you&#039;re mostly right

-3x^2+6x+12=0

-3(x^2-2x-4)=0    (up to this point is right)

then divide both sides by the -3

so you get

x^2-2x-4=0

[i don&#039;t think you could factor it more]

if it was -3x^2+6x+12=4, you would be right in making

x^2-2x-4=-4/3

you could also move it over so it&#039;s

x^2-2x-(8/3)

also, you can make the negative go away b/c you&#039;re dividing both sides by -1, and zero divided by anything will always equal zero

as long as you keep your signs straight, you&#039;ll be fine

that&#039;s as much as i can help you, good luck!

Answer 2

common factoring negative out

say you have -x^2-x+6&gt;0 you have to common factor - rite but since you divide everything in left side by -1 you have to divide everything on right side by negative one as well? so -(x^2+x-6)&lt;0 and -(x-2)(x+3) &lt;0 rite

Almost: When you divide by -1, you have to drop that negative sign.

You are here:

-1(x² + x - 6) &gt; 0 ← Divide both sides by -1 AND flip your sign.

x² + x - 6 &lt; 0

(x + 3)(x - 2) &lt; 0

x + 3 &lt; 0 and x - 2 &lt; 0

x &lt; -3 and x &lt; 2

x &lt; 2, x not equal to -3, will satisfy the equation.

***

second question -3x^2+6x+12=0

so would it be -3(x^2-2x-4)=0 ← It would be this...see below.

or if it was -3x^2+6x+12=4

it would be -3(x^2-2x-4)=4/(-3)?

-3(x² - 2x - 4) = 0

You will need to use the quadratic formula to solve from here.

x = [-b ± √(b² - 4ac)] / 2a; where a = 1, b = -2, c = -4

x = [-(-2) ± √({-2}² - 4*1*(-4))] / 2*1

x = [2 ± √(4 + 16)] / 2

x = [2 ± √20] / 2

x = [2 ± 2√5] / 2

x = 1 ± √5

Answer 3

common factoring negative out

say you have -x^2-x+6&gt;0 you have to common factor - rite but since you divide everything in left side by -1 you have to divide everything on right side by negative one as well? so -(x^2+x-6)&lt;0 and -(x-2)(x+3) &lt;0 rite

WRONG

when you multiply or divide an inequality by a negative you must reverse the direction of the inequality symbol, so you get

x^2 + x - 6 &lt; 0

you just factored out -1 on the left side, so the symbol should not be reversed in that step.

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second question -3x^2+6x+12=0

so would it be -3(x^2-2x-4)=0

YES now divide both sides by -3 and get x^2-2x-4=0

factor (x-4)(x+2)=0

x=4

x=-2

or if it was -3x^2+6x+12=4

it would be -3(x^2-2x-4)=4/(-3)?

NO it would be -3(x^2-2x-4)=4

but to solve this, first subtract 4 and get -3x^2+6x+8=0, then factor

    * 9 minutes ago

    * - 3 days left to answer.

Additional Details

9 minutes ago

are these correct

5 minutes ago

-x^2 - 3ix + 5 = 0

x^2 + 3ix - 5 = 0 how does this happen shouldnt ie be -(x^2+3ix+5)=0 how can the negative dissapear

to make the -1 &#039;disappear&#039; I would multiply everything by -1

x^2 +3ix -5=0