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2x(x2 + 3x - 7)
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∙ 9y ago
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Factor 4x5 plus 6x3 plus 6x2 plus 9?
Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)
What is the f dashed x or deriative of 1 over 2xcubed plus 5?
f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this
How do you factor 6x2 - 20x plus 2x3?
2x(x + 5)(x - 2)
What is the factorization of 3x3 plus 6x2 - 12x?
3x(x2 + 2x - 4)
Factor the trinomial 6x2 plus 14x -12?
(3x - 2)(2x + 6) or 2(3x - 2)(x + 3)
How do you list all factors of 6x2 plus 14x-12?
6x2 + 14x -12 = (3x - 2) (2x + 6). There they are.
What is the factorization of the polynomial 2x3 - 6x2 - 56x?
Step 1: 2x.(x2 - 3x - 28) Step 2: 2x.(x - 7)(x + 4)
Factor 4x5 plus 6x3 plus 6x2 plus 9?
Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)
Which is equivalent to 3x divide by 7 plus 5y divided by 14x?
3x/7 + 5y/14x = 3x*2x/(7*2x) + 5y/14x = (6x2 + 5y)/14x
What is the f dashed x or deriative of 1 over 2xcubed plus 5?
f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this
How do you factor 6x2 - 20x plus 2x3?
2x(x + 5)(x - 2)
What is 3x plus 1 times 2x plus 4?
(3x+1)(2x+4) = 6x2+4+2x+12x = 6x2+14x+4
Factor the trinomial 6x2 - 13x - 63?
6x2 - 13x - 63 = 6x2 + 14x - 27x - 63 = 2x(3x + 7) - 9(3x + 7) = (3x + 7)(2x - 9)
What is the factorization to 3x squared minus 14x minus 24?
3x2 -14x -24 can be factored as (3x + 4)(x - 6).
What is the factorization of 3x3 plus 6x2 - 12x?
3x(x2 + 2x - 4)
Factor the trinomial 6x2 plus 14x -12?
(3x - 2)(2x + 6) or 2(3x - 2)(x + 3)
What is factorization of 6x2-12x?
6x2 - 12x [take out what both terms have in common] 6x(x - 2) [if you multiply 6x by both terms in the parenthesis you will get your original answer]
