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A trinomial is an algebraic equation consisting of three terms. If the number of terms in an equation exceeds three, then it is commonly called polynomials. A quadratic trinomial has three terms in the equation, with the highest degree of the equation being 2. The standard form of a quadratic trinomial equation is ax² + bx + c. Factoring trinomials of the form x² + bx + c where c is negative is mostly similar to normally factoring a quadratic trinomial. Factorisation of a quadratic equation results in two root values for the equation that gives the values of the variable ‘x’.
Factoring trinomials
Factoring a trinomial refers to solving the equation.
When the given quadratic trinomial is solved, it results in two values.
These two values are the values of the variable ‘x’ in the equation.
There happen to be two values for ‘x’ because it is a quadratic equation.
A quadratic equation always has the highest degree of the equation as 2. Thus, there will be two roots to the equation.
Methods for solving trinomials
A quadratic trinomial is solved to find the root values. It can be done by three different methods.
Factoring
Quadratic formula
Completing the square
Any of the methods mentioned above give the values of the variable in the equation.
Steps for factoring trinomials
The factor is nothing but the multiples of the given number. When those two multiples are multiplied, it gives the considered product. The factors of a quadratic equation are nothing but the values of the variable in the equation. These values, when substituted in the equation, in the place of the variable, solve the equation to zero. These root values, when plotted in the graph, give a parabolic curve.
The common steps to factorise quadratic trinomials are:
Split the middle term.
Take the common factor out.
Take the common factor out again.
Equate the factors to zero.
To verify whether the root values are correct, substitute the values in the place of x separately, and see if the equation solves to zero.
Factoring trinomials of the form x² + bx + c, where c is negative
Trinomial in the form of x² + bx + c, where c is negative, means that the given quadratic equation is in the form ax² + bx – c. The only step that has a difference here is the splitting up of the middle term; that, too, the difference is only in terms of the signs of the split-up numbers.
Let’s consider the quadratic trinomial x² + 3x – 10 to solve by factoring.
The given quadratic trinomial is in the form ax² + bx + c, where c is negative.
That is, of the form ax² + bx – c.
1. Split the middle term.
The given equation is, x² + 3x – 10
The middle term 3x is split into 5x and -2x.
So, the sum of the two numbers gives the middle term, and their product equals the product of the coefficient of x² and the constant.
The important key to note is that, since the constant c is negative, one of the two numbers should be negative too.
Only then their product will be negative, as will be the product of the coefficient of x² and the constant.
Also, the bigger number should be positive since ‘b’ is positive.
Only when the bigger number is positive, on adding those two numbers, their sum will be positive and, so will be the middle term.
Thus, +3x is split into +5x and -2x.
x² + 3x – 10
= x² – 2x + 5x – 10
2. Take the common factor out.
Take out the common factors, such that
The first two terms are grouped under one parenthesis and
The last two terms are grouped separately under another one.
= x² – 2x + 5x – 10
= 1x(x – 2) + 5(x – 2)
3. Take the common factor out again.
Note that the terms inside the parenthesis are the same.
Take them out as common factors and group them separately.
= 1x(x – 2) + 5(x + 2)
= (x – 2) (1x + 5)
4. Equate the factors to zero.
Equate the factors in the two parentheses separately to zero.
It results in the values of ‘x’.
(x – 2) (1x + 5)
Equating (x – 2) to zero,
x – 2 = 0
x = 2
Equating (1x + 5) to zero,
1x + 5 = 0
1x = -5
x = -51
x = -5
To verify if the roots are correct, substitute them separately in the place of ‘x’ in the given equation.
Substituting x = 2 in the quadratic trinomial x² + 3x – 10,
x² + 3x – 10
= (2)² + 3(2) – 10
= 4 + 6 – 10
= 10 – 10
= 0. Verified to be correct.
Substituting x = -5 in the quadratic trinomial x² + 3x – 10,
x² + 3x – 10
= (-5)² + 3(-5) – 10
= 25 – 15 – 10
= 25 – 25
= 0. Verified to be correct.
Solved example for factoring trinomial of the form x² + bx + c, where c is negative
x² + 5x – 6
Splitting the middle term,
= x² – x + 6x – 6
Taking common factors out,
= x(x – 1) + 6(x – 1)
Taking common factors out,
= (x – 1) (x + 6)
Equating (x – 1) to zero,
x – 1 = 0
x = +1
Equating (x + 6) to zero,
x + 6 = 0
x = -6
Answer: x = 1, -6.
Conclusion
Factoring trinomials of the form x² + bx + c, where c is negative, is no different than the normal process. The only step to keep in mind is splitting up the middle term. While splitting the middle term into two, make sure that one among the two numbers is negative. If the middle term, b, is positive, then the bigger number is signed positive. The other number is negative. Whereas if b is negative, then the bigger number is negative, and the smaller number is positive.
