Mathematics, 31.08.2019 04:50, mjwaple57

Expand the following using the binomial theorem and pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 in the expansion of (3a + 4b)8, which of the following are possible variable terms? explain your reasoning. a2b3; a5b3; ab8; b8; a4b4; a8; ab7; a6b5

Answers: 1

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Answers

$\bf (2x+3)^5\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^5(+3)^0\\
2&+5&(2x)^4(+3)^1\\
3&+10&(2x)^3(+3)^2\\
4&+10&(2x)^2(+3)^3\\
5&+5&(2x)^1(+3)^4\\
6&+1&(2x)^0(+3)^5
\end{array}$

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element? well, 5*4/2
how did we get 10 for the fourth element? well, 10*2/4

$\bf (2x-3y)^4\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^4(-3y)^0\\
2&+4&(2x)^3(-3y)^1\\
3&+6&(2x)^2(-3y)^2\\
4&+4&(2x)^1(-3y)^3\\
5&+1&(2x)^0(-3y)^4
\end{array}$

$\bf (3a+4b)^8\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(3a)^8(+4b)^0\\
2&+8&(3a)^7(+4b)^1\\
3&+28&(3a)^6(+4b)^2\\
4&+56&(3a)^5(+4b)^3\\
5&+70&(3a)^4(+4b)^4\\
6&+56&(3a)^3(+4b)^5\\
7&+28&(3a)^2(+4b)^6\\
8&+8&(3a)^1(+4b)^7\\
9&+1&(3a)^0(+4b)^8
\end{array}$

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶

answered: jaidaarmstrong

The possible terms are a^5b^3,b^8,a^4b^4,a^8,ab^7 .

Step-by-step explanation:

Binomial Theorem:

$(a+b)^n=^nC_0a^n+^nC_1a^{n-1}b+...+^nC_{n-1}a^1b^{n-1}+^nC_nb^n$

Using Binomial Theorem, we get

$(x + 2)^6=^6C_0a^n+^6C_1a^{5}b+...+^6C_{5}a^1b^{5}+^5C_5b^5$

(x + 2)^6=x^6 + 12 x^5 + 60 x^4 + 160 x^3 + 240 x^2 + 192 x + 64

Similarly,

(x-4)^4=x^4 - 16 x^3 + 96 x^2 - 256 x + 256

(2x+3)^5=32 x^5 + 240 x^4 + 720 x^3 + 1080 x^2 + 810 x + 243

(2x-3y)^4=16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4

The Pascal’s triangle is given below. In Pascal’s triangle nth row represents the coefficients of the expression of $(a+n)^{n-1}$ .

In the expression (x + 2)^6 , a=x, b=2 and n=6. Using 7th row of Pascal’s triangle we get

(x + 2)^6=1(x^6)(2^0)+6(x^5)(2^1)+15(x^4)(2^2)+20(x^3)(2^3)+15(x^2)(2^4)+6(x^1)(2^5)+1(x^0)(2^6)

(x + 2)^6=x^6 + 12 x^5 + 60 x^4 + 160 x^3 + 240 x^2 + 192 x + 64

Similarly,

In (x-4)^4 a=x, b=-4 and n=4. Using 5th row of Pascal’s triangle we get

(x-4)^4=x^4 - 16 x^3 + 96 x^2 - 256 x + 256

In (2x+3)^5 a=2x, b=3 and n=5. Using 6th row of Pascal’s triangle we get

(2x+3)^5=32 x^5 + 240 x^4 + 720 x^3 + 1080 x^2 + 810 x + 243

In (2x-3y)^4 a=2x, b=-3y and n=4. Using 5th row of Pascal’s triangle we get

(2x-3y)^4=16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4

In binomial expansion of (3a + 4b)^8 the degree of each term must be 8. It means the sum of powers of a and b is equal to 8.

For a^2b^3 the dgree of term is 2+3=5. So, it is not possible variable term.

Therefore the possible terms are a^5b^3,b^8,a^4b^4,a^8,ab^7 .

Expand the following using the binomial theorem and pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (

answered: Guest

answer: idk

step-by-step explanation:

answered: Guest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)

0 = x (add numbers together on both sides)

so in order for the equation to be true, x must be 0.

trying x = -0.5

5 + x - 12 = 2x - 7 (original equation)

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)

5 + -0.5 - 12 = -1 - 7 (did multiplication)

-7.5 = -8 (did addition on both sides)

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.

trying x = 0

5 + x - 12 = 2x - 7 (original equation)

5 + 0 - 12 = 2(0) - 7 (plug in value of x)

5 + 0 - 12 = 0 - 7 (did multiplication)

-7 = -7 (did addition on both sides)

since -7 equals -7, that demonstrates that the value of 0 for x is correct.

trying x = 1

5 + x - 12 = 2x - 7 (original equation)

5 + 1 - 12 = 2(1) - 7 (plug in value of x)

5 + 1 - 12 = 2 - 7 (did multiplication)

-6 = -5 (did addition on both sides)

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

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Expand the following using the binomial theorem and pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (...