# Expanding (x + a)(x +b)(x + c)

In this chapter, we will learn to expand the algebraic term (x + a)(x + b)(x + c) and derive a shortcut formula for fast calculation.

At the end of the chapter, we have also provided solved problems for your practice.

## Formula for (x + a)(x + b)(x + c)

To derive the formula, we have to multiply and expand the expression.

Let’s first multiply the two terms.

\mathtt{\Longrightarrow \ ( x+a)( x+b)( x+c)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} +xb+ax+ab\right)( x+c)}

Now multiply the terms again.

\mathtt{\Longrightarrow \ x^{3} +bx^{2} +ax^{2} +abx+cx^{2} +bcx+acx+abc}\\\ \\ \mathtt{\Longrightarrow \ x^{3} +ax^{2} +bx^{2} +cx^{2} +abx+bcx+acx+abc}\\\ \\ \mathtt{\Longrightarrow \ x^{3} +( a+b+c) x^{2} +( ab+bc+ca) x+abc}

Hence, the formula for (x + a)(x + b)(x + c) can be expressed as;

\mathtt{\ x^{3}} + (Sum of all coefficient) \mathtt{\ x^{2}} + (Sum of product of two coefficient).x + (product of all coefficient)

You have to memorize this formula as it would help you solve questions faster in the examinations.

I hope you understood the concept. Let us solve some problems for better clarity.

### (x + a)(x +b)(x + c) – Solved problems

Example 01
Simplify the expression (x + 3)(x + 9)(x + 2)

Solution
Referring the formula;

⟹ (x + a)(x + b)(x + c)

\mathtt{\Longrightarrow \ x^{3} +( a+b+c) x^{2} +( ab+bc+ca) x+abc}

Where;
a = 3
b = 9
c = 2

Putting the values in formula, we get;

\mathtt{\Longrightarrow \ x^{3} +( 3+9+2) x^{2} +( 3.9\ +\ 9.2+2.3) x+( 3)( 9)( 2)}\\\ \\ \mathtt{\Longrightarrow \ \ x^{3} +14x^{2} +( 27\ +18+6) x+54}\\\ \\ \mathtt{\Longrightarrow \ x^{3} +14x^{2} +51x+54}

Hence, above expression is the solution.

Example 02
Expand the below expression
(x – 1)(x – 3)(x + 4)

Solution
Referring the formula;

⟹ (x + a)(x + b)(x + c)

\mathtt{\Longrightarrow \ x^{3} +( a+b+c) x^{2} +( ab+bc+ca) x+abc}

Where;
a = -1
b = -3
c = 4

Putting the values, we get;

\mathtt{\Longrightarrow \ x^{3} +( -1-3+4) x^{2} +(( -1)( -3) +( -3) .4+4( -1)) x+( -1)( -3)( 4)}\\\ \\ \mathtt{\Longrightarrow \ \ x^{3} +0.x^{2} +( 3-12-4) x+12}\\\ \\ \mathtt{\Longrightarrow \ x^{3} -13x+12}

Hence, the above expression is the solution.

Example 03
Solve the below expression
(x + 1)(x + 2)(x – 7)

Solution
Referring the formula;

⟹ (x + a)(x + b)(x + c)

\mathtt{\Longrightarrow \ x^{3} +( a+b+c) x^{2} +( ab+bc+ca) x+abc}

Where;
a = 1
b = 2
c = -7

Putting the values, we get;

\mathtt{\Longrightarrow \ x^{3} +( 1+2-7) x^{2} +( 1.2+2.( -7) +( -7) .1) x+( 1)( 2)( -7)}\\\ \\ \mathtt{\Longrightarrow \ \ x^{3} -4x^{2} +( 2-14-7) x-14}\\\ \\ \mathtt{\Longrightarrow \ x^{3} -4x^{2} -19x-14}

Hence, the above expression is the solution.

Example 04
Solve the below expression.
(3x + 1) (3x + 5)(3x + 6)

Solution
Notice that in the place of x, we have 3x here. So, in the formula, we will replace x with 3x.

Referring the formula;

⟹ (x + a)(x + b)(x + c)

\mathtt{\Longrightarrow \ x^{3} +( a+b+c) x^{2} +( ab+bc+ca) x+abc}

Here;
a = 1
b = 5
c = 6

Putting the values;

\mathtt{\Longrightarrow \ ( 3x)^{3} +( 1+5+6)( 3x)^{2} +( 1.5+5.6+6.1)( 3x) +( 1)( 5)( 6)}\\\ \\ \mathtt{\Longrightarrow \ \ 27x^{3} +12\ \left( 9x^{2}\right) \ +( 5+30+6)( 3x) +30}\\\ \\ \mathtt{\Longrightarrow \ 27x^{3} +108x^{2} +123x+30}

Hence, the above expression is the solution.

Next chapter : Algebraic identity grade 09 problems

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