What is (x+1) to the 3rd power? how did you solve it?

Question

need help on this homework. not just the answer but how you did it. if it was squared i know you foil it. PLZ HELP. and if your good at math, snend me your aim if you'd love to help me lol

Answer 1

    (x + 1)^3 = x^3 + 3x^2(1) + 3x(1)^2 + (1)^3

    (x+1)^3 = x^3 + 3x^2 + 3x + 1

Answer 2

That's the problem with "tricks" like "foil." They only work in a very few instances and are useless most of the time.

(x+1)^3 = [(x+1)(x+1)](x+1)

= [(x+1)(x) + (x+1)(1)](x+1) because of the distributive property of multiplication over addition: a(b+c)=ab+ac

[(x+1)(x) + (x+1)(1)](x+1) = [(x)(x+1) + (1)(x+1)](x+1) because of the commutative property of multiplication: ab = ba

[(x)(x+1) + (1)(x+1)](x+1) = [(x)(x) + (x)(1) + (1)(x) + (1)(1)](x+1) because of that distributive property again.

[(x)(x) + (x)(1) + (1)(x) + (1)(1)](x+1) = [x^2 + 2x + 1](x+1)

[x^2 + 2x + 1](x+1) = [x^2 + 2x + 1](x) + [x^2 + 2x + 1](1) that distributive thing again.

[x^2 + 2x + 1](x) + [x^2 + 2x + 1](1) = [(x^2)(x) + (2x)(x) + (1)(x)] + [(x^2)(1) + (2x)(1) + (1)(1)] that distributivity thing again.

And that = x^3 + 3x^2 + 3x + 1 after you multiply out those terms,and do some addition.

You're going to hear a lot of "tricks." Do yourself a BIG favor and forget them.

Remember:

Commutative Property of Addition: a+b=b+a

Commutative Property of Multiplication: ab=ba

Associative Property of Addition (a+b)+c=a+(b+c)

Associative Property of Multiplication (ab)c=a(bc)

Distributive Property of Multiplication over Addition: a(b+c)=ab+ac

Property of 0 a+0=0+a=a

Property of 1. a1=1a=a

Special Property of 0: a0=0a=0

Additive Inverses: For any number a you give me, I can give you a number (its Additive Inverse) so that when you add your number and mine, you get 0 (a)+(-a) = 0

Multiplicative Inverses: For ALMOST any number you give me, I can give you a number so that when you multiply your number and mine, you get 1. (a)(1/a)=1

One number doesn't have a Multiplicative Inverse. Remember that Special Property of 0? Zero multiplied by anything gives you zero. There is no "1/0" such that (0)(1/0)=1. People call multiplying by a multiplicative inverse, "dividing by the number." And that's why they tell you "You can't divide by zero."

THESE RULES ALWAYS WORK... No tricks. No shortcuts. You'll have enough stuff to memorize without having to memorize a bunch of tricks that only work in a very few instances.

Answer 3

(x + 1)³ =

(x + 1) (x + 1) (x + 1) =

(x² + x + x + 1) (x + 1) =

(x² + 2x + 1) (x + 1) =

x³ + x² + 2x² + 2x + x + 1 =

x³ + 3x² + 3x + 1

Answer 4

= (x + 1)³

= (x + 1)(x + 1)(x + 1)

= (x² + x + x + 1)(x + 1)

= (x² + 2x + 1)(x + 1)

= x³ + 2x² + x + x² + 2x + 1

= x³ + 3x² + 3x + 1

Answer: x³ + 3x² + 3x + 1

Answer 5

same philosophy, just with one, you're going to have 3 elements, just multiply everything in one parens with everything in the other.

(x+1)(x+1)(x+1)

(x^2 + 2x + 1)(x+1)

x^3 + x^2 + 2x^2 + 2x + x +1

x^3 + 3x^2 + 3x + 1

Answer 6

You can foil this also.

You would have

(x+1)^3 = (x+1) * (x+1) * (x+1)

Multiply the first x+1 quantities together to get

(x+1)^3  =  (x^2 + x + x + 1) * (x+1)

              =  (x^2 + 2x + 1) * (x+1).

Then multiplying  the two above quantities you get this

(x +1)^3 = x^3 + x^2 + 2x^2 + 2x + x  + 1

The answer simplified is

(x + 1)^3   =  x^3 + 3x^2 + 3x + 1.

Note: Alternative method below if it is too confusing the distributive method will always work.

You can also use the binomial theorem

For the left most term in the binomial, raise it to the highest power and decrease it for each successive term.

x^3 + x^2 + x^1 + x^0

For the right most term in the binomial, raise it to the lowest power and increase it for each successive term.

1^0 + 1^1 + 1^2 + 1^3

Multiply the corresponding leftmost and rightmost binomial terms

x^3 * x^0 + x^2 *1^1 + x^1*1^2 + x^0*1^3 .

The final step is to multiply by the binomial coefficients which you can get from the combinations formula

The combination formula is

               n C r   =     (n!) / [(n - r!) * r ! ]

Since the power is 3 in this case

n = 3 and r starts off at zero and then increases by 1 from each term left to right.

(x + 1)^3 = 3 C 0 [x^3 * x^0] + 3 C 1 [x^2 *1^1]

                   + 3 C 2[ x^1*1^2] + 3 C 3 [ x^0*1^3 ].

The above equals

(x+1)^3    x^3 + 3x^2 + 3x + 1

The coefficients in the binomial come from the combinations formula.

You can also get binomial coefficients using Pascal's triangle

          1    0 th row

         11    1st row

         1 2 1  2nd row

        1 3 3 1   3rd row    (row you want for (x+1)^3

        1 4 6 4 1

        1 5 10 5 1

Answer 7

(x+1)^3=(x+1)^2(x+1)

(x+1)^2=x^2+2x+1

(x^2+2x+1)(x+1)

just multiply everything in left by everything on right.

x^3+2x^2+x+x^2+2x+1

x^3+3x^2+3x+1

make it a good day