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What are the common factors of 5x plus 10 and 6x2 plus 12x?
5x + 10 = 5(x+2) 6x2 + 12x = 6x(x+2) The only common factor is x+2
What is the length of a rectangle when the area is 6x2 plus 19x plus 10 and the width is 3x plus 2?
6x2+19x+10 = (2x+5)(3x+2) Length = 2x+5 Width = 3x+2
What are the x-solutions for the equation 10x2-64 equals 36 plus 6x2?
If: 10x2-64 = 36+6x2 Then: 4x2-100 = 0 And: (2x-10)(2x+10) = 0 So: x = 5 or x = -5
How do you factor six x squared plus sixty four x minus twelve equals ten?
6x2 + 64x - 12 = 10 ∴ 6x2 + 64x - 22 = 0 ∴ 3x2 + 32x - 11 = 0 ∴ 3x2 + 33x - x - 11 = 0 ∴ 3x(x + 11) - 1(x + 11) = 0 ∴ (3x - 1)(x + 11) = 0 ∴ x ∈ {1/3, -11}
How do you factor x2-20x plus 100?
(x - 10)(x - 10)
What are the common factors of 5x plus 10 and 6x2 plus 12x?
5x + 10 = 5(x+2) 6x2 + 12x = 6x(x+2) The only common factor is x+2
2x4 plus 6x2-10 how do you put this in a quadratic form?
10
What is the length of a rectangle when the area is 6x2 plus 19x plus 10 and the width is 3x plus 2?
6x2+19x+10 = (2x+5)(3x+2) Length = 2x+5 Width = 3x+2
Factor 6x2 plus 11x-10?
6x2 + 11x - 10 = (6x2 + 15x) - (4x + 10) = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5). Note, first, that the three co-efficients of the original quadratic expression are 6, 11, and -10. We need two numbers whose sum is 11 and whose product is (6)(-10) = -60. These two numbers turn out to be 15 and -4. Thus, we replace the middle term, 11x with 15x - 4x and proceed with the factorisation in the usual way.
What are the x-solutions for the equation 10x2-64 equals 36 plus 6x2?
If: 10x2-64 = 36+6x2 Then: 4x2-100 = 0 And: (2x-10)(2x+10) = 0 So: x = 5 or x = -5
What does 4-2-6x2?
= -10
What is the factor trinomial 2x plus 7x plus 10?
10 + 9x
Factor x2 plus 7x plus 10?
1
How do you factor six x squared plus sixty four x minus twelve equals ten?
6x2 + 64x - 12 = 10 ∴ 6x2 + 64x - 22 = 0 ∴ 3x2 + 32x - 11 = 0 ∴ 3x2 + 33x - x - 11 = 0 ∴ 3x(x + 11) - 1(x + 11) = 0 ∴ (3x - 1)(x + 11) = 0 ∴ x ∈ {1/3, -11}
What is the solution set for 6x2-x-15 equals 0?
6x2 - x - 15 = 0 since (6)(-15) = (-10)(9) and -10 + 9 = -1, then factor as: [(6x + 9)/3][6x - 10)/2] = 0 (2x + 3)(3x - 5) = 0 let each factor be zero: 2x + 3 = 0 or 3x - 5 = 0 solve for x: The solutions are: x = -3/2 and x = 5/3
Where does the graph of y 6x2 11x minus 10 cross the x axis?
If: 6x2+11x-10 = 0 Then: x = -5/2 and x = 2/3
How do you factor 10c plus 10?
10c + 10The only common factor is 10, and this can be divided out, resulting in:10(c+1)
