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When the given expression equals 0 then x = -1/6 and x = -6

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6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.

This equation can't actually be factored, but you can solve for x: 6x2 + 37x + 6 = 0 ∴ 6x2 + 37x = -6 ∴ x2 + (37/6)x = -1 ∴ x2 + (37/6)x + (37/12)2 = -1 - (37/12)2 ∴ (x + 37/12)2 = -1 - (37/12)2 ∴ x + 37/12 = ±[-1 - (37/12)2]1/2 ∴ x = 37/12 ±[-1 - (37/12)2]1/2 Which gives you two complex numbers for the answer, approximately: 37 / 12 + 3.2414417 × i 37 / 12 - 3.2414417 × i

(6 × 2) + 3 = 15

6x2-18x+12 = (6x-6)(x-2)

6x2 + 14x -12 = (3x - 2) (2x + 6). There they are.

6x2 + 10x = 5 6x2 + 10x - 5 = 0 The roots are 1/(2*6)* [-10 +/- sqrt(102 - 4*6*(-5))] = 1/(2*6) * [-10 +/- sqrt 220] =1/6 * [-5 +/- sqrt 55] So x = -2.0694 or 0.403

34+(6*2)=34+12+46

x5+4x4-6x2+nx+2 when divided by x+2 has a remainder of 6 Using the remainder theorem: n = 2

6(x - 2)(x - 1)

x3 + 6x2 - 4x - 24 = (x + 6)(x2 - 4) = (x + 6)(x + 2)(x - 2)

It is: 0

5 + 6xÂ² + 3x - 7xÂ³ + x6 would have the degree of 6.

6x2 + 5x - 6 is factored out to (2x + 3)(3x - 2).

3/(x2+4x-1) = 6/(2x2-3x+5) Cross - multiply in order to eliminate the fractions: 6*(x2+4x-1) = 3*(2x2-3x+5) 6x2+24x-6 = 6x2-9x+15 6x2-6x2+24x+9x = 15+6 33x = 21 x = 21/33 => x = 7/11

I do not believe that there are any rational roots.

2x(3x+6) = 0 x = 0 or x = -2

It is: 12-4(6*-1) = 25

Divide all terms by 3 and it is: (2x-1)(x+2)

6 multiplied by 4 is 24 here is how i did it 6x1=6 6x2=12 (you just do 6 plus another 6) 6x3=18 6x4=24 hope this helped :)

x2 + 11x + 30 = 0 (x + 5)(x + 6) = 0 so the roots are -5 and -6

First divide by 6: 6(x2 - 3x + 2) = 6(x - 1)(x - 2)

6x2= 12.

(3x - 2)(2x + 6) or 2(3x - 2)(x + 3)

x = 2.236, -2.236 & -6. (to 3 decimal places)x3 + 6x2 - 5x - 30 = 0First of all, let us note that as the highest power in the polynomial is to the power of 3, we are expecting to find 3 roots (the real + imaginary roots will equal 3).Ok, so we need to factor the expression x3 + 6x2 - 5x - 30.There are different methods to try to factor an expression like this. We are going to split this down into 2 groups (because we can see that we will be able to factor each).1.x3 + 6x2 &2. -5x - 30. We can factor out the "-" (minus) here to become: - (5x + 30).So we have:(x3 + 6x2) - (5x + 30) = 0We can factor out "x2" from the first bracket and "5" from the second to get:x2 (x + 6) -5 (x + 6) = 0We can now see that "x + 6" is a common factor and so we can rewrite the original expression as:(x2 - 5) (x + 6) = 0Now we can solve for the roots of x by making each bracket equal to 0 in turn, so:x2 - 5 = 0x2 = 5x = 51/2 (we have two roots here of course: the square root of 5 & the square root of 5 times -1)x = 2.236 & -2.236 (to 3 decimal places only)&x + 6 = 0x = -6Thus all roots are real (none are imaginary) and they are: x = 2.236, -2.236 & -6.

6 * x2 = 6x2