Precalculus Examples

${(x + 4 y)}^{3}$

Step 1

Pascal's Triangle can be displayed as such:

$1$

$1 - 1$

$1 - 2 - 1$

$1 - 3 - 3 - 1$

The triangle can be used to calculate the coefficients of the expansion of ${(a + b)}^{n}$ by taking the exponent $n$ and adding $1$ . The coefficients will correspond with line $n + 1$ of the triangle. For ${(x + 4 y)}^{3}$ , $n = 3$ so the coefficients of the expansion will correspond with line $4$ .

Step 2

The expansion follows the rule ${(a + b)}^{n} = c_{0} a^{n} b^{0} + c_{1} a^{n - 1} b^{1} + c_{n - 1} a^{1} b^{n - 1} + c_{n} a^{0} b^{n}$ . The values of the coefficients, from the triangle, are $1 - 3 - 3 - 1$ .

$1 a^{3} b^{0} + 3 a^{2} b + 3 a b^{2} + 1 a^{0} b^{3}$

Step 3

Substitute the actual values of $a$ $x$ and $b$ $4 y$ into the expression.

$1 {(x)}^{3} {(4 y)}^{0} + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4

Simplify each term.

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Step 4.1

Multiply ${(x)}^{3}$ by $1$ .

${(x)}^{3} {(4 y)}^{0} + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.2

Apply the product rule to $4 y$ .

$x^{3} (4^{0} y^{0}) + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.3

Rewrite using the commutative property of multiplication.

$4^{0} x^{3} y^{0} + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.4

Anything raised to $0$ is $1$ .

$1 x^{3} y^{0} + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.5

Multiply $x^{3}$ by $1$ .

$x^{3} y^{0} + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.6

Anything raised to $0$ is $1$ .

$x^{3} \cdot 1 + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.7

Multiply $x^{3}$ by $1$ .

$x^{3} + 3 {(x)}^{2} {(4 y)}^{1} + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.8

Simplify.

$x^{3} + 3 x^{2} (4 y) + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.9

Rewrite using the commutative property of multiplication.

$x^{3} + 3 \cdot 4 x^{2} y + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.10

Multiply $3$ by $4$ .

$x^{3} + 12 x^{2} y + 3 {(x)}^{1} {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.11

Simplify.

$x^{3} + 12 x^{2} y + 3 x {(4 y)}^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.12

Apply the product rule to $4 y$ .

$x^{3} + 12 x^{2} y + 3 x (4^{2} y^{2}) + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.13

Rewrite using the commutative property of multiplication.

$x^{3} + 12 x^{2} y + 3 \cdot 4^{2} x y^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.14

Raise $4$ to the power of $2$ .

$x^{3} + 12 x^{2} y + 3 \cdot 16 x y^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.15

Multiply $3$ by $16$ .

$x^{3} + 12 x^{2} y + 48 x y^{2} + 1 {(x)}^{0} {(4 y)}^{3}$

Step 4.16

Multiply ${(x)}^{0}$ by $1$ .

$x^{3} + 12 x^{2} y + 48 x y^{2} + {(x)}^{0} {(4 y)}^{3}$

Step 4.17

Anything raised to $0$ is $1$ .

$x^{3} + 12 x^{2} y + 48 x y^{2} + 1 {(4 y)}^{3}$

Step 4.18

Multiply ${(4 y)}^{3}$ by $1$ .

$x^{3} + 12 x^{2} y + 48 x y^{2} + {(4 y)}^{3}$

Step 4.19

Apply the product rule to $4 y$ .

$x^{3} + 12 x^{2} y + 48 x y^{2} + 4^{3} y^{3}$

Step 4.20

Raise $4$ to the power of $3$ .

$x^{3} + 12 x^{2} y + 48 x y^{2} + 64 y^{3}$