Expansion of (x + a)(x + b)(x + c): The key to understanding the concept of Expansion of Powers of Binomials and Trinomials is through brackets. You can understand how to multiply the binomial expression with the help of brackets. In this article you can find quick and easy methods for Expansion of (x + a)(x + b)(x + c). Scroll down this page to know the derivation of Expansion of (x + a)(x + b)(x + c) with suitable examples.
Expansion of (x + a)(x + b)(x + c)
Here we will discuss the expansion of (x + a)(x + b)(x + c) with the derivation with explanation.
(x + a)(x + b)(x + c) = (x + a){(x + b)(x + c)}
(x + a)(x + b)(x + c) = x {(x + b)(x + c)} + a {(x + b)(x + c)}
(x + a)(x + b)(x + c) = x {x(x + c) + b(x + c)} + a {x(x + c) + b(x + c)}
(x + a)(x + b)(x + c) = x {x² + cx + bx + bc} + a {x² + cx + bx + bc}
(x + a)(x + b)(x + c) = x³ + cx² + bx² + bcx + ax² + acx + abx + abc
(x + a)(x + b)(x + c) = x³ + x²(a + b + c) + x(ab + bc + ac) + abc
Thus the expansion of (x + a)(x + b)(x + c) is x³ + x²(a + b + c) + x(ab + bc + ac) + abc
This can be read as cubed x + squared x {sum of the three constant terms} + x{sum of the product of the three constant terms} + product of three constant terms.
Do Refer:
- Expansion of (x ± a)(x ± b)
- Expansion of (a ± b ± c)^2
- Worksheet on Expansion of (a ± b)^2 and its Corollaries
Expansion of (x + a)(x + b)(x + c) Examples
Step by step explanation for each and every question is provided here.
Example 1.
Find the product of (x + 1) (x + 3) (x + 6) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x + 1) (x + 3) (x + 6)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = 1, b = 3, c = 6
Therefore
(x + 1) (x + 3) (x + 6) = x³ + (1 + 3 + 6)x² + (1 × 3 + 3 × 6 + 6 × 1)x + 1× 3× 6
(x + 1) (x + 3) (x + 6) = x³ + 10x² + 27x + 18
Thus the product of (x + 1) (x + 3) (x + 6) is x³ + 10x² + 27x + 18
Example 2.
Find the product of (x + 2) (x – 1) (x -3) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x + 2) (x – 1) (x – 3)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = 1, b = 3, c = 6
Therefore
(x + 2) (x – 1) (x – 3) = x³ + (2-1-3)x² + (2(-1)+(-1)(-3)+(-3)(2))x + 2(-1)(-3)
(x + 2) (x – 1) (x – 3) = x³ + (-2)x² + (-2+3-6)x + 6
(x + 2) (x – 1) (x – 3) = x³ – 2x² + 7x + 6
Thus the product of (x + 2) (x – 1) (x – 3) is x³ – 2x² + 7x + 6
Example 3.
Find the product of (x – 1) (x – 2) (x – 3) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x – 1) (x – 2) (x – 3)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = -1, b = -2, c = -3
Therefore
(x – 1) (x – 2) (x – 3) = x³ + (-1-2-3)x² + ((-1)(-2)+(-2)(-3)+(-3)(-1))x + (-1)(-2)(-3)
(x – 1) (x – 2) (x – 3) = x³ – 6x² + 11x – 6
Thus the product of (x – 1) (x – 2) (x – 3) is x³ – 6x² + 11x – 6
Example 4.
Find the product of (x + 3) (x +4) (x +6) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x + 3) (x + 4) (x + 6)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = 3, b = 4, c = 6
Therefore
(x + 3) (x + 4) (x + 6) = x³ + (3 + 4 + 6)x² + (3×4 + 4×6 6×3)x + 3×4×6
(x + 3) (x + 4) (x + 6) = x³ + 13x² + 54x + 72
Thus the product of (x + 3) (x + 4) (x + 6) is x³ + 13x² + 54x + 72
Example 5.
Find the product of (x – 1) (x + 2) (x + 4) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x – 1) (x + 2) (x + 4)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = -1, b = 3, c = 6
Therefore
(x – 1) (x + 2) (x + 4) = x³ + (-1 + 3 + 6)x² + (-1(3) + 3 × 6 + 6(-1))x + (-1) × 3 × 6
(x – 1) (x + 2) (x + 4) = x³ + 8x² + 9x – 18
Thus the product of (x – 1) (x + 2) (x + 4) is x³ + 8x² + 9x – 18