While factorizing trinomials of the form x^{2} + ax + b, where a and b are integers, we have to express the trinomial as a product of two binomials.

The trinomial x^{2} + ax + b can be expressed as the product of two binomials of the form (x - A)(x - B), such that, A + B = a and AB = b.

Consider x^{2} + 4x + 3. This is a trinomial of the form x^{2} + ax + b where, a = 4 and b = 3.

x^{2} + 4x + 3 = (x + 1)(x + 3)

Observe that 3 + 1 = 4, which is the coefficient of x and 3 x 1 = 3 which is the constant term.

Let us consider another trinomial x^{2} - x - 6. This trinomial can be factored as (x + 2)(x - 3). Here, the sum of 2 and -3 is -1, which is the coefficient of x. The product of 2 and -3 is -6, which is the constant term.

While factoring trinomials of the form x^{2} + ax + b into binomials of the form (x + A)(x + B),

- The sum of the constant terms of a binomial must be equal to the coefficient of x in the trinomial i.e. A + B = a
- The product of the constant terms in the binomial must be equal to the constant term in the trinomial i.e AB = b
- The sign of the constant term in the binomial is the same as the sign of the coefficient of x in the trinomial if the constant term in the trinomial is positive. i.e if b > 0, then A and B have the same sign as a.
- If the constant term in the trinomial is negative, then the constant term in the binomial have opposite signs. i.e. if b is negative, then A and B have the opposite signs.

We can find the factors that satisfy all these conditions in a methodic way. Write down all the factors of the constant term in the trinomial and write down the sum of the factors. For example, consider the trinomial x^{2} + 5x - 24.

The factors of the constant term -24 and the sum of the factors are tabulated as follows:

** Factors ** |
** Sum of Factors ** |

2, -12 |
2 + (-12) = -10 |

-2, 12 |
-2 + 12 = 10 |

3, -8 |
3 + (-8) = -5 |

-3, 8 |
-3 + 8 = 5 |

4, -6 |
4 + (-6) = -2 |

-4, 6 |
-4 + 6 = 2 |

-3 and 8 are two constants such that -3 x 8 is -24 and -3 + 8 is 5. The trinomial x^{2} + 5x - 24 = (x - 3)(x + 8)

Similarly, the trinomial of the form ax

^{2} + bx + c, where a, b and c are

real numbers and a ≠ 0 can be factored as a product of binomials with two more points to be taken care of

- If the trinomial does not have any common factors, then the binomials will not have any common factors as well.
- List the factors of a and c and then list all the combinations of the factors and check for the products.

When the leading coefficient is one, the factors can be formed by splitting the middle coefficient as the factors of the last term and pairing each with the variable present.

### Examples of Factoring Trinomials

Given below are some of the examples that explain how to factorize trinomials.

**Example 1: **

Factorize 3x^{3} + 24x^{2} + 45x

**Solution: **

First find the GCF of the terms in the given trinomial. The GCF of the terms 3x^{3}, 24x^{2}, 45x is 3x

Now, factor the GCF out of all the terms.

3x^{3} + 24x^{2} + 45x = 3x(x^{2}) + 3x(8x) + 3x(15)

= 3x(x^{2} + 8x + 15)

Now, factor the trinomial x

^{2} + 8x + 15

** Factors ** |
** Sum ** |

15, 1 |
16 |

3, 5 |
8 |

The sum of the factors 3 and 5 is 8. Hence the trinomial, x^{2} + 8x + 15 = (x + 3)(x + 5)

3x^{3} + 24x^{2} + 45x = 3x(x^{2} + 8x + 15)

= 3x(x+3)(x+5)

Hence, the trinomial 3x^{3} + 24x^{2} + 45x is factored to 3x(x+3)(x+5)

**Example 2:**

Factorize the trinomial 4y^{2} - 27y + 18.

**Solution: **

4y^{2} - 27y + 18 is a trinomial that does not have any common factors. And so, the binomial factors also do not have any common factors.

The constant term of the trinomial is +18. It is positive and so the signs of the constant terms of the binomial factors have the same sign. The coefficient of y is -27. It is negative and so the constant terms of the binomial factors have a negative sign.

** Factors of 4 ** |
** Factors of 18 ** |

1, 4 |
-1, -18 |

2, 2 |
-2, -9 |

4, 1 |
-3, -6 |

Using different combinations of these factors, let us write the binomial factors

(y - 1)(4y - 18) - There is a common factor

(y-2)(4y - 9)

(y - 3)(4y - 6) - There is a common factor

(2y - 1)(2y - 18) - There is a common factor

(2y - 2)(2y - 9) - There is a common factor

(2y - 3)(2y - 6) - There is a common factor

(4y - 1)(y -18)

(4y - 2)(y - 9) - There is a common factor

(4y - 3)(y - 6)

Now, we are left with three combinations

(y - 2)(4y - 9)

(4y - 1)(y - 18)

(4y - 3)(y - 6)

Calculate the middle terms for each of them

The middle term of (y - 2)(4y - 9) is -8y - 9y = -17y

The middle term of (4y - 1)(y - 18) is -72y - y = -73y

The middle term of (4y - 3)(y - 6) is -3y - 24y = -27y

Therefore, 4y^{2} - 27y + 18 = (4y - 3)(y - 6)