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3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =
(12 + 15 + x2 -5)/x2 =
(27 - 5 + x2)/x2 =
(22 + x2)/x2
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∙ 12y ago
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What is the answer for x2 plus 1 plus 3x4 equals 2?
-7.5?
What is the remainder when 2x4 plus 5x3 - 17x2 - 5 is divided by x plus 3?
Quotient: 2x3-x2-14x+42 Remainder: -131 over (x+3)
How do you solve x2 plus 3x4?
This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========
Simplify x2-7x plus 6 divided x2-36?
i got 42 divided 7x
What is the expntents of x5 - 3x4 plus x2 - 4 plus 5x4 plus 3x2 plus 2x plus 1?
x^5+2x^4+4x^2+2x-3
Quotient and remainder of 3X4 plus 2X3 minus X2 minus x minus 6 divided by X2 plus 1?
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
What is the answer for x2 plus 1 plus 3x4 equals 2?
-7.5?
What is the remainder when 2x4 plus 5x3 - 17x2 - 5 is divided by x plus 3?
Quotient: 2x3-x2-14x+42 Remainder: -131 over (x+3)
Factor 5x3 plus 40y6?
5(x + 2y2)(x2 - 2xy2 + 4y4)
How do you solve x2 plus 3x4?
This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========
-6x3 plus 4x7-0.8x2 plus x4-5x5?
-6x3 + 4x7 - 0.8x2 + x4 - 5x5 = 4x7 - 5x5 + x4 - 6x3 - 0.8x2 = x2 (4x5 - 5x3 + x2 - 6x - 0.8)
What is the coefficient of the term of degree 4 in this polynomial 2x5 plus 3x4 - x3 plus x2 - 12?
it is 3. You are doing APEX right?
How do you find the missing term in a problem like for example 5X3 plus 5X equals 5X times?
Since 5x is a factor of both terms, divide it. 5x3 + 5x = 5x(x2 + 1)
Simplify x2-7x plus 6 divided x2-36?
i got 42 divided 7x
What is the expntents of x5 - 3x4 plus x2 - 4 plus 5x4 plus 3x2 plus 2x plus 1?
x^5+2x^4+4x^2+2x-3
Given 3x3 plus 4x2 plus x plus 7 is divided by x2 plus 1?
Given 3x3 + 4x2 +x + 7 is divided by x2 + 1, find the results:
X6 plus 3x4-x2-3 equals 0?
x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.
