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divide by 3: 3(6x2 + 13x - 5)
Look for factors of 6 which can be multiplied by factors of -5 and added to give 13.
Factors of 6: 1 & 6; 2 & 3;
Factors of -5: -1 & 5; -5 & 1
the last and first ones give 3 x -5 = -15 and 2 x 1 = 2, total 13 so:
3(3x - 1)(2x + 5) = 3(6x2 - 2x + 15x - 5) = 18x2 + 39x - 15
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How do you factor the trinomial completely 18x3 29x 3x?
I'm guessing that's 18x3 + 29x2 + 3x. That factors to x(2x + 3)(9x + 1) 18x3 + 29x2 + 3x factor x = x(18x2 + 29x + 3) write 29x as 27x + 2x = x(18x2 + 27x + 2x + 3) group by two and factor each group = x[(18x2 + 27x) + (2x + 3)] = x[9x(2x + 3) + 1(2x + 3)] continue factoring = x(2x + 3)(9x + 1) Or since (18)(3) = 54 and (27)(2) = 54 then write: = x(18x2 + 29x + 3) = x[(18x + 27)/9][(18x + 2)/2] = x(2x + 3)(9x + 1)
Can you factor x squared plus 4x plus 2?
No
Factor out the Greatest Common Factor of 4a plus 8?
The greatest common factor (GCF) is 4.
What is the GCF factor of 8m plus 8n?
The GCF is 8. 8(m + n)
What is the factor to 4ac plus 2ad plus 2bc plus bd?
(2a + b)(2c + d)
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Factor 3x3 - 18x2 plus 24x?
3x(x - 2)(x - 4)
How do you factor 9x3 plus 18x2 - x - 2?
(x + 2)(3x - 1)(3x + 1)
Is 18x2 a factor of 36?
That's a factor string of 36.
How do you factor 18x2-2?
2(3x + 1)(3x - 1)
How do you factor 18x2 - 2?
2(3x - 1)(3x + 1)
X3 - 18x2 plus 17x?
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What is the factorization of this trinomial 3x3 - 18x2 plus 24x?
3x3-18x2+24x=3x(x2-6x+8)=3x(x-4)(x-2)
How do you factor -3x2-18x-24?
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What is the square root of 18x squared plus 12x plus two?
18x2 +12x + 2 this is not a perfect square trinomial so let's work a little bit here. 18x2 +12x + 2 factor 2 = 2(9x2 +6x + 1) = 2(3x + 1)2 represent 2 as (√2)2 = (√2)2(3x + 1)2 = [(√2)(3x + 1)]2 take its square root = (√2)(3x + 1) for all x > -1/3.
How do you factor 18X squared plus 2?
18x2 + 2= 2(9x2 + 1)= 2(9x2 - -1)= 2[(3x)2 - i2]= 2(3x + √-1)(3x - √-1) (substitute i for √-1)= 2(3x + i)(3x - i)
How do you factor the trinomial completely 18x3 29x 3x?
I'm guessing that's 18x3 + 29x2 + 3x. That factors to x(2x + 3)(9x + 1) 18x3 + 29x2 + 3x factor x = x(18x2 + 29x + 3) write 29x as 27x + 2x = x(18x2 + 27x + 2x + 3) group by two and factor each group = x[(18x2 + 27x) + (2x + 3)] = x[9x(2x + 3) + 1(2x + 3)] continue factoring = x(2x + 3)(9x + 1) Or since (18)(3) = 54 and (27)(2) = 54 then write: = x(18x2 + 29x + 3) = x[(18x + 27)/9][(18x + 2)/2] = x(2x + 3)(9x + 1)
