3x2 + 22x + 7 = 3x2 +21x + x + 7 = 3x(x + 7) + (x + 7) = (x + 7)(3x + 1)
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
no 3x2=6 then 4x-15= -60 so 6 added with - 60 = -54
3x^2=-20 so x^2=-20/3 this has no real roots but the plus of minus i(square root of 20/3) is a solution if we allow complex numbers.
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
3x2 + 22x + 7 = 3x2 +21x + x + 7 = 3x(x + 7) + (x + 7) = (x + 7)(3x + 1)
x=-.25.
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
3x2=6+38=44
In general, no.
This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========
x3+3x2-22x-90 divided by x-5 equals x2+8x+18 For the answer to be correct then the quotient multiplied by the divisor must be equal to the dividend. Hence:- (x-5)*(x2+8x+18) Multiplying out the brackets term by term: x3-5x2+8x2-40x+18x-90 Collecting like terms in descending order: x3+3x2-22x-90 So the answer is correct.
y2=x3+3x2
3x2 + x - 2 = 0(3x - 2) (x + 1) = 0x = + 2/3x = -1
No real roots.
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}