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it is 87.

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Q: What is 29x3?
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Can you show me the factor tree of 87?

29x3 both prime so your done


Is 87 a prime or composite?

It has more than 2 factors; 1, 29, and 87. 29x3=87. Therefore, it is composite.


What number comes next in the sequence 22-104-268-427?

One possible answer is 494. This uses the cubic function (-29x3 + 256x2 - 401x + 218)/2


2x-3 parenthesis 2x plus 3 parenthesis equals?

(2x-3)(2x+3) can be multiplied out using the FOIL method, which stands for:FirstOuterInnerLastThis method only works for the multiplication of two binomials, so don't rely exclusively upon it.For the "first" part, multiply the two first terms of the binomials together:(2x)(2x)=4x2For the "outer" part, multiply the two outermost terms together:(2x)(3)=6xFor the "inner" part, mulitply the two innermost terms together:(2x)(-3)=-6xFor the "last" part, mulitply the two last terms of each binomial together:(3)(-3)=-9Add all of these sub-calculations together to get your final result:4x2+6x-6x-9As you can see, 6x-6x=0, so these two terms will cancel out, leaving:4x2-9Another way of doing this multiplication is to see it as an application of the distributive property, but instead of a single number distributed across a binomial, it is another binomial. This is what the FOIL method essentially does, but it creates a handy mnemonic to remember it. To see it as a distributive problem, visualize it like this as I distribute (2x-3) across (2x+3):(2x-3)(2x+3)=(2x-3)(2x)+(2x-3)(3)=(4x2-6x)+(6x-9)=4x2-9As you can see, the same answer is reached. This method of distributing one entire term across the other polynomial holds true for more complicated multiplications, so it is the most accurate way to memorize how to handle these problems. A simplification of something like this:(x-7)(5x4+4x3-x2+9x-2)cannot be handled with such a convenient crutch as the FOIL method, but realizing it is simply distribution makes it very doable:(x-7)(5x4)+(x-7)(4x3)-(x-7)(x2)+(x-7)(9x)-(x-7)(2)(5x5-35x4)+(4x4-28x3)-(x3-7x2)+(9x2-63x)-(2x-14)5x5-35x4+4x4-28x3-x3+7x2+9x2-63x-2x+145x5-31x4-29x3+16x2-65x+14And in case that side-track example distracted you from the real answer, it was 4x2-9


Related questions

Can you show me the factor tree of 87?

29x3 both prime so your done


Is 87 a prime or composite?

It has more than 2 factors; 1, 29, and 87. 29x3=87. Therefore, it is composite.


What number comes next in the sequence 22-104-268-427?

One possible answer is 494. This uses the cubic function (-29x3 + 256x2 - 401x + 218)/2


2x-3 parenthesis 2x plus 3 parenthesis equals?

(2x-3)(2x+3) can be multiplied out using the FOIL method, which stands for:FirstOuterInnerLastThis method only works for the multiplication of two binomials, so don't rely exclusively upon it.For the "first" part, multiply the two first terms of the binomials together:(2x)(2x)=4x2For the "outer" part, multiply the two outermost terms together:(2x)(3)=6xFor the "inner" part, mulitply the two innermost terms together:(2x)(-3)=-6xFor the "last" part, mulitply the two last terms of each binomial together:(3)(-3)=-9Add all of these sub-calculations together to get your final result:4x2+6x-6x-9As you can see, 6x-6x=0, so these two terms will cancel out, leaving:4x2-9Another way of doing this multiplication is to see it as an application of the distributive property, but instead of a single number distributed across a binomial, it is another binomial. This is what the FOIL method essentially does, but it creates a handy mnemonic to remember it. To see it as a distributive problem, visualize it like this as I distribute (2x-3) across (2x+3):(2x-3)(2x+3)=(2x-3)(2x)+(2x-3)(3)=(4x2-6x)+(6x-9)=4x2-9As you can see, the same answer is reached. This method of distributing one entire term across the other polynomial holds true for more complicated multiplications, so it is the most accurate way to memorize how to handle these problems. A simplification of something like this:(x-7)(5x4+4x3-x2+9x-2)cannot be handled with such a convenient crutch as the FOIL method, but realizing it is simply distribution makes it very doable:(x-7)(5x4)+(x-7)(4x3)-(x-7)(x2)+(x-7)(9x)-(x-7)(2)(5x5-35x4)+(4x4-28x3)-(x3-7x2)+(9x2-63x)-(2x-14)5x5-35x4+4x4-28x3-x3+7x2+9x2-63x-2x+145x5-31x4-29x3+16x2-65x+14And in case that side-track example distracted you from the real answer, it was 4x2-9