Pre-Algebra Examples

$(4 x - 7) (5 x + 27) = 20 x^{2} - 27 - 14$

Step 1

Simplify $(4 x - 7) (5 x + 27)$ .

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Step 1.1

Expand $(4 x - 7) (5 x + 27)$ using the FOIL Method.

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Step 1.1.1

Apply the distributive property.

$4 x (5 x + 27) - 7 (5 x + 27) = 20 x^{2} - 27 - 14$

Step 1.1.2

Apply the distributive property.

$4 x (5 x) + 4 x \cdot 27 - 7 (5 x + 27) = 20 x^{2} - 27 - 14$

Step 1.1.3

Apply the distributive property.

$4 x (5 x) + 4 x \cdot 27 - 7 (5 x) - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2

Simplify and combine like terms.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Rewrite using the commutative property of multiplication.

$4 \cdot 5 x \cdot x + 4 x \cdot 27 - 7 (5 x) - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.1.2.1

Move $x$ .

$4 \cdot 5 (x \cdot x) + 4 x \cdot 27 - 7 (5 x) - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2.1.2.2

Multiply $x$ by $x$ .

$4 \cdot 5 x^{2} + 4 x \cdot 27 - 7 (5 x) - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2.1.3

Multiply $4$ by $5$ .

$20 x^{2} + 4 x \cdot 27 - 7 (5 x) - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2.1.4

Multiply $27$ by $4$ .

$20 x^{2} + 108 x - 7 (5 x) - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2.1.5

Multiply $5$ by $- 7$ .

$20 x^{2} + 108 x - 35 x - 7 \cdot 27 = 20 x^{2} - 27 - 14$

Step 1.2.1.6

Multiply $- 7$ by $27$ .

$20 x^{2} + 108 x - 35 x - 189 = 20 x^{2} - 27 - 14$

Step 1.2.2

Subtract $35 x$ from $108 x$ .

$20 x^{2} + 73 x - 189 = 20 x^{2} - 27 - 14$

Step 2

Subtract $14$ from $- 27$ .

$20 x^{2} + 73 x - 189 = 20 x^{2} - 41$

Step 3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx 2.02739726$

Step 4