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Q: What is the factorization to 19x cubed minus 38x squared?

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x(x - 19)(x - 1)

x2 - 19x + 60 = x2 - 15x - 4x + 60 = x(x - 15) - 4(x - 15) = (x - 15)(x - 4)

3x2 - 19x - 14 = 0 3x2 - 21x + 2x - 14 = 0 3x(x - 7) + 2(x - 7) = 0 (x - 7)(3x + 2) = 0 x - 7 = 0 or 3x + 2 = 0 x = 7 or x = -2/3

(x - 12)(x - 7)

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Factor as : (x+9)(x+10) using the first terms ( x terms) multiply to x2; the last terms multiply to 90; the sum of 9x + 10 x = 19x

-11

-19x/38xy = -1x0/y = -1/y

simplified it would equal: 24x to the third power - x to the second power - x to the sixth power - 34

8x - 7 = 15 + 19xsubtract 19x from both sidesadd seven to both sides8x -19x = 15 +7-11x = 22divide both sides by -11x = -2

19x = 437 x = 437 ÷ 19 x = 23

x^2-19x+90 (x-9) (x-10)

19x

3

152

19x 100

19x(10^3)

19x+1x+37=0 First add the 19x and 1x together 20x+37=0 Next subtract 37 to the other side 20x=-37 Now divide both sides by 20 x=-20/37

Let x = number of teachers 19x = number of students x + 19x = 500 20x = 500 x = 25 teachers 19x = 475 students

(x - 10)(x - 9)

+9+9=19x

27x - 5y

(7x + 2)(x - 3)

have changed your equation from 19x=29 mod16 to 19x=16 mod29 as it is impossible to have 29 mod 16 in modular arithmetic/it makes no sense. Anyway, any modular equation a=b mod m can be rewritten as a-b=km where k is an integer. or in other words, m divides a-b without remainder. Applying this to the equation above we get (19x-16)/29=k. Now just stick it into a calculator and replace x with integers until the equation gives out an integer, in this case x turns out to be 10.

In a complex expression two (or more) factors are combined with one another by multiplication. Terms are combined by addition (or subtraction). For example, in the expression (2x - 3)(3x - 5), the factors are (2x - 3) and (3x - 5).If you multiply it out, you get 6x2 - 19x + 15 where the terms are 6x2, 19x and 15. You could consider -19x as a term instead and then all are additive.

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