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X3 plus x2 minus 6x plus 4?
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
How do you solve x3 plus 4x2 plus 6x plus 24?
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
How do you factor x3 plus 5x2 minus 16x minus 80?
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
Find factors of x3 plus 12x2 plus 48x plus 64?
x3 + 12x2 + 48x + 64= (x + 4)(x2 + 8x + 16)= (x + 4)(x + 4)(x + 4)= (x + 4)3
How do you factor x3-3x2 plus 4?
x3 - 3x2 + 4 Since the coefficients of the odd powers of x (=1) is the same as the sum of the even powers (-3+4=1), then x = -1 must be a root. That is to say, (x + 1) is a factor. So you can rewrite the expression as x3 + x2 - 4x2 - 4x + 4x + 4 = x2(x + 1) - 4(x + 1) + 4(x + 1) = (x + 1)*(x2 - 4x + 4) = (x + 1)*(x - 2)2
What is the factor of x3 plus x2 plus 4x plus 4?
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
Factor x3 plus 4x2 plus x plus 4?
x3 + 4x2 + x + 4 = (x + 4)(x2 + 1)
How do you find a cubic equation with integral coefficients that has the roots 2 and 4 plus i?
If it has integral coefficients and 4+i is a root then its conjugate, 4-i must also be a root. So the equation is f(x) = (x-2)*(x-4-i)*(x-4+i) where each factor is x minus a root. Then multiply these out. = (x-2)*(x2 - 8x + 17) = x3 - 2x2 - 8x2 + 16x + 17x - 34 = x3 - 10x2 + 33x - 34
X3 plus x2 minus 6x plus 4?
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
How do you solve x3 plus 4x2 plus 6x plus 24?
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
How do you factor x3 plus 5x2 minus 16x minus 80?
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
Find factors of x3 plus 12x2 plus 48x plus 64?
x3 + 12x2 + 48x + 64= (x + 4)(x2 + 8x + 16)= (x + 4)(x + 4)(x + 4)= (x + 4)3
Is x3 plus 4 a trinomial?
no it is not. it is only a binomial.
The coefficients in the expansion of (x plus y)4 are?
1,4,6,4,1
What is the quotient and remainder if any when the expession 4x4 -x3 plus 17x2 plus 11x plus 4 is divided by 4x plus 3?
Dividend: 4x4-x3+17x2+11x+4 Divisor: 4x+3 Quotient: x3-x2+5x-1 Remainder: 7
What is x3 plus x plus 1 divided by x-1?
4
How do you factor x3-3x2 plus 4?
x3 - 3x2 + 4 Since the coefficients of the odd powers of x (=1) is the same as the sum of the even powers (-3+4=1), then x = -1 must be a root. That is to say, (x + 1) is a factor. So you can rewrite the expression as x3 + x2 - 4x2 - 4x + 4x + 4 = x2(x + 1) - 4(x + 1) + 4(x + 1) = (x + 1)*(x2 - 4x + 4) = (x + 1)*(x - 2)2
