MULTIPLYING BINOMIAL BY TRINOMIAL

Expand and simplify :

Example 1 :

(x + 2)(x² + x + 4)

Solution :

By distributing x and 2, we get

= x(x²) + x(x) + x(4) + 2(x²) + 2(x) + 2(4)

= x³ + x² + 4x + 2x² + 2x + 8

= x³ + 3x² + 6x + 8

Example 2 :

(x + 3)(x² + 2x - 3)

Solution :

By distributing x and 3, we get

= (x + 3)(x² + 2x - 3)

= x(x²) + x(2x) - x(3) + 3(x²) + 3(2x) - 3(3)

= x³ + 2x² - 3x + 3x² + 6x - 9

= x³ + 5x² + 3x - 9

Example 3 :

(x + 3)(x² + 2x + 1)

Solution :

By distributing x and 3, we get

= (x + 3)(x² + 2x + 1)

= x(x²) + x(2x) + x(1) + 3(x²) + 3(2x) + 3(1)

= x³ + 2x² + x + 3x² + 6x + 3

= x³ + 5x² + 7x + 3

Example 4 :

(x + 1)(2x² - x - 5)

Solution :

By distributing x and 1, we get

= (x + 1)(2x² - x - 5)

= x(2x²) - x(x) - x(5) + 1(2x²) - 1(x) - 1(5)

= 2x³ - x² - 5x + 2x² - x - 5

= 2x³ + x² - 6x - 5

Example 5 :

(2x + 3)(x² + 2x + 1)

Solution :

By distributing 2x and 3, we get

= (2x + 3)(x² + 2x + 1)

= 2x(x²) + 2x(2x) + 2x(1) + 3(x²) + 3(2x) + 3(1)

= 2x³ + 4x² + 2x + 3x² + 6x + 3

= 2x³ + 7x² + 8x + 3

Example 6 :

(2x - 5)(x² - 2x - 3)

Solution :

By distributing 2x and 5, we get

= (2x - 5)(x² - 2x - 3)

= 2x(x²) - 2x(2x) - 2x(3) - 5(x²) + 5(2x) + 5(3)

= 2x³ - 4x² - 6x - 5x² + 10x + 15

= 2x³ - 9x² + 4x + 15

Example 7 :

(x + 5)(3x² - x + 4)

Solution :

By distributing x and 5, we get

= (x + 5)(3x² - x + 4)

= x(3x²) - x(x) + x(4) + 5(3x²) - 5(x) + 5(4)

= 3x³ - x² + 4x + 15x² - 5x + 20

= 3x³ + 14x² - x + 20

Example 8 :

(4x - 1)(2x² - 3x + 1)

Solution :

By distributing 4x and 1, we get

= (4x - 1)(2x² - 3x + 1)

= 4x(2x²) - 4x(3x) + 4x(1) - 1(2x²) + 1(3x) - 1(1)

= 8x³ - 12x² + 4x - 2x² + 3x - 1

= 8x³ - 14x² + 7x - 1