Question

Use pascal's triangle to expand the binomial (d-5y)^6 show work.

Answer 1

(d - 5y)⁶ = d⁶ - 30d⁵y + 375d⁴y² - 2500d³y³ + 9375d²y⁴ - 18750dy⁵ + 15625y⁶

Further explanation

The Problem:

Use Pascal's triangle to expand the binomial  (d - 5y)⁶.

The Process:

Look carefully at Pascal's triangle scheme in the attached picture.

Let us do a binomial expansion to:

$\boxed{ \ (a - b)^6 = a^6 - 6a^5b + 15a^4b^2 - 20a^3b^3 + 15a^2b^4 - 6ab^5 + b^6 \ }$, which comes from the following processing:

$\boxed{ \ (a - b)^6 = a^6 + 6a^5(-b) + 15a^4(-b)^2 + 20a^3(-b)^3 + 15a^2(-b)^4 - 6a(-b)^5 + (-b)^6 \ }$

Alright, see carefully how the expansion of this binomial expression.

$\boxed{ \ (d - 5y)^6 = ? \ }$

$\boxed{ \ (d - 5y)^6 = d^6 - 6d^5(5y) + 15d^4(5y)^2 - 20d^3(5y)^3 + 15d^2(5y)^4 - 6d(5y)^5 + (5y)^6 \ }$

$= d^6 - (5 \times 6)d^5y + (25 \times 15)d^4y^2 - (125 \times 20)d^3y^3 + (625 \times 15)d^2y^4 - (3125 \times 6)dy^5 + 15625y^6$

Thus, the expansion form of the binomial expression (d - 5y) ⁶ is $\boxed{\boxed{ \ d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6 \ }}$

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Alternative Steps

We can also use Newton's Binomial Expansion.

$\boxed{ \ (a + b)^n = \sum_{r = 0}^{n} \left(\begin{array}{ccc}n\\r\end{array}\right) a^{n- r}b^r \ }$

Remember this $\boxed{ \ \left(\begin{array}{ccc}n\\r\end{array}\right) = \frac{n!}{r!(n-r)!} \ }$

$\boxed{ \ (d - 5y)^6 = ? \ }$

$\boxed{ \ \sum_{r = 0}^{6} \left(\begin{array}{ccc}6\\0\end{array}\right) d^{6-0}(-5y)^6 \ }$

$\boxed{ \ = \left(\begin{array}{ccc}6\\0\end{array}\right) d^6(-5y)^0 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\1\end{array}\right) d^5(-5y)^1 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\2\end{array}\right) d^4(-5y)^2 \ }$ +  $\boxed{ \ \left(\begin{array}{ccc}6\\3\end{array}\right) d^3(-5y)^3 \ }$ +  $\boxed{ \ \left(\begin{array}{ccc}6\\4\end{array}\right) d^2(-5y)^4 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\5\end{array}\right) d^1(-5y)^5 \ }$ + $\boxed{ \ \left(\begin{array}{ccc}6\\6\end{array}\right) d^0(-5y)^6 \ }$

$\boxed{ \ = d^6 - 6d^5(5y) + 15d^4(25y^2) - 20d^3(125y^3) + 15d^2(625y^4) - 6d(3125y^5) + 15625y^6 \ }$

$\boxed{\boxed{ \ (d - 5y)^6 = d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6 \ }}$

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Notes

Pascal's triangle in common is a triangular array of binomial coefficients. It is undoubtedly entitled after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in Italy, Germany, Persia, India, and China.

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Another example

$\boxed{ \ (2x - \frac{1}{2x})^3 = ? \ }$

$\boxed{ \ = 1 \cdot (2x)^3 + 3 \cdot (2x)^2 \Big(-\frac{1}{2x} \Big) + 3 \cdot 2x \Big(-\frac{1}{2x} \Big)^2 + 1 \cdot \Big(-\frac{1}{2x} \Big)^3 \ }$

$\boxed{ \ = (2x)^3 - 3 \cdot (2x)^2 \Big(\frac{1}{2x} \Big) + 3 \cdot 2x \Big(\frac{1}{2x} \Big)^2 - \Big(\frac{1}{2x} \Big)^3 \ }$

$\boxed{ \ = 8x^3 - 12x^2 \Big(\frac{1}{2x} \Big) + 6x \Big(\frac{1}{4x^2} \Big) - \Big(\frac{1}{8x^3} \Big) \ }$

$\boxed{ \ 8x^3 - 6x + \frac{3}{2x} - \frac{1}{8x^3} \ } \rightarrow \boxed{ \ = 8x^3 - 6x + \frac{3}{2}x^{-1} - \frac{1}{8}x^{-3} \ }$

• 2nd term in expansion of $\boxed{ \ (2x - \frac{1}{2x})^3 \ }$ is -6x.
• The coefficient of x⁻¹ in expansion of $\boxed{ \ (2x - \frac{1}{2x})^3 \ }$ is ³/₂.

Learn more

1. Which expression is equivalent to the product of a binomial and a trinomial after it has been fully simplified? brainly.com/question/1394854
2. A polynomial of the 5th degree with a leading coefficient of 7 and a constant term of 6 brainly.com/question/12700460
3. Which binomial is not a divisor of x³ - 11x² + 16x + 84? brainly.com/question/5694594

Answer 2

In attached file you can see Pascal's triangle. Explanation for this triangle is following:

1. coefficient in 0th row is $(a+b)^0=1$
2. coefficients in 1st row are coefficients of $(a+b)^1=1\cdot a+1\cdot b$
3. coefficients in 2nd row are coefficients of $(a+b)^2=1\cdot a^2+2\cdot ab+1\cdot b^2$
4. coefficients in 3rd row are coefficients of $(a+b)^3=1\cdot a^3+3\cdot a^2b+3\cdot ab^2+1\cdot b^3$ and so on
5. coefficients in 6th row are coefficients of $(a+b)^6=1\cdot a^6+6\cdot a^5b+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot ab^5+1\cdot b^6.$

Then

$(d-5y)^6=\\=1\cdot d^6-6\cdot d^5\cdot 5y+15\cdot d^4\cdot (5y)^2-20\cdot d^3\cdot (5y)^3+15\cdot d^2\cdot (5y)^4-6\cdot d\cdot (5y)^5+1\cdot (5y)^6.$

Simplify it:

$(d-5y)^6=d^6-30d^5y+375d^4y^2-2500d^3y^3+9375d^2y^4-18750dy^5+15625y^6.$