Solving Quadratic Equations by Factoring :
In this section, you will learn how to solve a quadratic equation by factoring.
Generally we have two types of quadratic equation.
The general form of a quadratic equation is
ax^{2} + bx + c = 0
In a quadratic equation, leading coefficient is nothing but the coefficient of x^{2}.
(i) In a quadratic equation in the form ax^{2} + bx + c = 0, if the leading coefficient is 1, we have to decompose the constant term "c" into two factors.
(ii) The product of the two factors must be equal to the constant term "c" and the addition of two factors must be equal to the coefficient of x, that is "b".
(iii) If p and q are the two factors of the constant term c, then we have to factor the quadratic equation using p and q as shown below.
(x + p)(x + q) = 0
(iv) Solving the above equation, we get
x = -p and x = -q
Quadratic Equation |
Signs of Factors |
ax^{2} + bx + c = 0 |
Positive sign for both the factors. |
ax^{2} - bx + c = 0 |
Negative sign for both the factors. |
ax^{2} + bx - c = 0 |
Negative sign for smaller factor and positive sign for larger factor. |
ax^{2} - bx - c = 0 |
Positive sign for smaller factor and negative sign for larger factor. |
Example :
Solve :
x^{2} + 17 x + 60 = 0
Solution :
The given quadratic equation is in the form of
ax^{2} + bx + c = 0
Check whether the coefficient of x^{2} is 1 or not.
Because the coefficient of x^{2} is 1, we have to decompose 60 into two factors as shown below.
Because the constant term 60 is having positive sign, both the factors must be positive.
In the above four pairs of factors, we have to select the a pair of factors such that the product of two factors is equal to the constant term "+60" and the addition of two factors is equal to the coefficient of x, that is "+17".
Now, factor the given quadratic equation and solve for x as shown below.
(x + 12)(x + 5) = 0
x + 12 = 0 or x + 5 = 0
x = -12 or x = -5
So, the solution is {-12, -5}.
(i) In a quadratic equation in the form ax^{2} + bx + c = 0, if the leading coefficient is not 1, we have to multiply the coefficient of x^{2} and the constant term. That is "ac". Then, decompose "ac" into two factors.
(ii) The product of the two factors must be equal to "ac" and the addition of two factors must be equal to the coefficient of x, that is "b".
(iii) Divide the two factors by the coefficient of x^{2} and simplify as much as possible.
(iv) Write the remaining number along with x (This is explained in the following example).
Example :
Solve :
2x^{2} + x - 6 = 0
Solution :
The given quadratic equation is in the form of
ax^{2} + bx + c = 0
Here, the coefficient of x^{2} is 1 or not.
Multiply the coefficient of x^{2} and the constant term "-6".
That is,
2 ⋅ (-6) = -12
Decompose -12 into two factors such that the product of two factors is equal to -12 and the addition of two factors is equal to the coefficient of x, that is 1.
Then, the two factors of -12 are
4 and -3
Now we have to divide the two factors 4 and -3 by the coefficient of x^{2}, that is 2.
Now, factor the given quadratic equation and solve for x as shown below.
(x + 2)(2x - 3) = 0
x + 2 = 0 or 2x - 3 = 0
x = -2 or x = 3/2
x = -2 or x = 1.5
So, the solution is {-2, 1.5}.
Problem 1 :
Solve the quadratic equation by factoring :
x^{2} – 5x – 24 = 0
Solution :
In the given quadratic equation, the coefficient of x^{2} is 1.
Decompose the constant term -24 into two factors such that the product of the two factors is equal to -24 and the addition of two factors is equal to the coefficient of x, that is 5.
Then, the two factors of -24 are
+3 and -8
Factor the given quadratic equation using and +3 and solve for x.
(x + 3)(x - 8) = 0
x + 3 = 0 or x - 8 = 0
x = -3 and x = 8
So, the solution is {-3, 8}.
Problem 2 :
Solve the quadratic equation by factoring :
3x^{2} – 5x – 12 = 0
Solution :
In the given quadratic equation, the coefficient of x^{2} is not 1.
So, multiply the coefficient of x^{2} and the constant term "-12".
3 ⋅ (-12) = -36
Decompose -36 into two factors such that the product of two factors is equal to -36 and the addition of two factors is equal to the coefficient of x, that is -5.
Then, the two factors of -36 are
+4 and -9
Now we have to divide the two factors 4 and -3 by the coefficient of x^{2}, that is 3.
Now, factor the given quadratic equation and solve for x as shown below.
(3x + 4)(x - 3) = 0
3x + 4 = 0 or x - 3 = 0
x = -4/3 or x = 3
So, the solution is {-4/3, 3}.
Problem 3 :
Solve the quadratic equation by factoring :
(x + 3)^{2} - 81 = 0
Solution :
Write the given quadratic equation in the form
ax^{2} + bx + c = 0
Then,
(x + 3)^{2} - 81 = 0
(x + 3)(x + 3) - 81 = 0
x^{2} + 3x + 3x + 9 - 81 = 0
x^{2} + 6x - 72 = 0
In the quadratic equation above, the coefficient of x^{2} is 1.
Decompose the constant term -72 into two factors such that the product of the two factors is equal to -72 and the addition of two factors is equal to the coefficient of x, that is +6.
Then, the two factors of -72 are
+12 and -6
Factor the given quadratic equation using +12 and -6 and solve for x.
(x + 12)(x - 6) = 0
x + 12 = 0 or x - 6 = 0
x = -12 and x = 6
So, the solution is {-12, 6}.
After having gone through the stuff given above, we hope that the students would have understood how to solve quadratic equations by factoring.
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