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Q: If x/3 + 2 = 6 then what is x?

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X3 - 4 = 2 X3 = 6 X = 61/3 X = about 1.817

x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)

x3 + 2x2 - 5x - 6 = x3 + x2 + x2 + x - 6x - 6 = x2(x + 1) + x(x + 1) - 6(x + 1) = (x + 1)(x2 + x - 6) = (x + 1)(x2 + 3x - 2x - 6) = (x + 1)[x(x + 3) - 2(x + 3)] = (x + 1)(x + 3)(x - 2)

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

x3 + 2x2 - 3x - 6 = (x2 - 3)(x + 2)

(x-2)(x^2+3)

The answer to x4+x3-14x2+4x+6 divided by x-3 is x3+4x2-2x-2

72 ---> 2 x 36 2x36 ---> 2 x 6 x6 2 x 6 x6 ---> 2 x 3 x 3x 3 -x3 good luck

-2

(x - 2)(x + 4)(x - 6)

Yes.

(x + 2)(x^2 + 2x + 3)

x3 - 4x2 + x + 6 The sum of the odd coefficients equals the sum of the even coefficients, so (x + 1) is a factor. So x3 - 4x2 + x + 6 = x3 + x2 - 5x2 - 5x + 6x + 6 = x2(x + 1) - 5x(x + 1) + 6(x + 1) = (x + 1)(x2 - 5x + 6) = (x + 1)(x - 2)(x - 3)

x3 - 4x2 + x + 6 = (x + 1)(x - 2)(x - 3)

x3-4=2 Add four to both sides. x3=6 Take the cu=6 be root of both sides. x is equal to the cube root of 6, or approximately 1.81712.

x3 - x2 + x - 2 has no rational factors.

x3 - 2x2 + x - 2 =(x - 2)(x2 + 1)

There is no factorisation for x3-2 nor for x^3-2.

x3 + 13x2 + 42x = x(x + 6)(x + 7).

(x+2)3 =(x)3+3(x)2(2)+3(x)(2)2+(2)3 =x3+6x2+12x+6

x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)

x4 + 3x3 - x2 - 9x - 6 = 0 x4 + x3 + 2x3 + 2x2 - 3x2 - 3x - 6x - 6 = 0 x3(x + 1) + 2x2(x + 1) - 3x(x + 1) - 6(x + 1) = 0 (x + 1)(x3 + 2x2 - 3x - 6) = 0 (x + 1)[x2(x + 2) - 3(x + 2)] = 0 (x + 1)(x + 2)(x2 - 3) = 0 So x + 1 = 0 so that x = -1 or x + 2 = 0 so that x = -2 or x2 - 3 = 0 so that x = +/- sqrt(3)

x3 + 8 = x3 + 23 = (x + 2)(x2 + 2x + 22) = (x + 2)(x2 + 2x + 4)

I know the answer is x(x^2+3x+6) + 4(x^2+3x+6) which factors to (x+4)(x^2+3x+6) but I don't know the rule behind this. Could someone explain it to me?

x3 + 4x2 + x - 6 = 0 x3 - x2 + 5x2 - 5x + 6x - 6 = 0 x2(x - 1) + 5x(x - 1) + 6(x - 1) = 0 (x - 1)(x2 + 5x + 6) = 0 (x - 1)(x + 2)(x + 3) = 0 So x = 1 or x = -2 or x = -3