Sometimes, quadratic equations can be solved by factorizing, which is also called grouping. The solve by factoring process is usually consists of three major steps:
All terms should be moved to one side of the equation using addition or subtraction. Rewrite the equation so that the left side of the equation is set equal to 0. For example, if the original equation is
, it should be rewritten as
.
![Rendered by QuickLaTeX.com x^2 = 5x - 3](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-a6c728b22b6001f2ef33fb62830bdb9e_l3.png)
![Rendered by QuickLaTeX.com x^2 - 5x + 3 = 0](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-fe8b2e9a306681b0a8a9b5af0935323a_l3.png)
The equation should be factored completely. It will have two factors.
Each factor should be set equal to zero. Since factors are of first power, they are easy to be solved.
All of the solutions should be combined to obtain the full solution set for the original equation.
The catch of this method is the finding of factors. Except some trivial cases, factoring quadratic equations is not easier than using other methods to solve the quadratic equations. Here are a few tips to help you factorize a quadratic equations:
1) Use polynomial identities:
![Rendered by QuickLaTeX.com (a+b)^2 = a^2 + 2ab + b^2](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-c5b3911ed013009f930bc93d5ef4b4fe_l3.png)
![Rendered by QuickLaTeX.com x^2 + (a+b)x + ab = (x + a)(x + b)](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-feeaf6b2ed20017ecd4b702b9755848a_l3.png)
![Rendered by QuickLaTeX.com a^2 - b^2 = (a+b)(a-b)](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-cf3bd1117c363a2c103172b61adab7f8_l3.png)
Example 1:
Solve the quadratic equation
.
![Rendered by QuickLaTeX.com x^2+2x+1=0](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-efff0fb6236c746727c7cbe8d15650ce_l3.png)
Comparing with
, we see that substituting
and
we have,
![Rendered by QuickLaTeX.com (a+b)^2 = a^2 + 2ab + b^2](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-fb4b68ade3796e8d49dcb239464be026_l3.png)
![Rendered by QuickLaTeX.com a=x](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-51fc686a349c3693edcf1669e7c72a9f_l3.png)
![Rendered by QuickLaTeX.com b=1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-611c476a70e4c711919f17fb0b692a9e_l3.png)
![Rendered by QuickLaTeX.com x^2+2x+1=0 \to (x+1)^2=0 \to (x+1)=0 \to x=-1.](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-dc4ec1da91d68cbe868793f53ac4a1da_l3.png)
Example 2:
Solve the quadratic equation
.
![Rendered by QuickLaTeX.com y^2+3y+2=0](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-f89aa9fea4e2c64c5ab0e5f0dd44b19c_l3.png)
To use the identity
we need to find
and
such that their summation is
and multiplication equals to latex
. It is not that hard to “guess” that
and
satisfy the condition. Therefore,
![Rendered by QuickLaTeX.com x^2 + (a+b)x + ab = (x + a)(x + b)](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-fefa00bd0a4087346d3b5f232cdff46b_l3.png)
![Rendered by QuickLaTeX.com a](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-244c5aa5a58f308f9f7b3b4ffcd88a40_l3.png)
![Rendered by QuickLaTeX.com b](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-8f9d5549ee12f468d8090e0656e97500_l3.png)
![Rendered by QuickLaTeX.com 3](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-51ce3209903725750a8093b4c262dffd_l3.png)
![Rendered by QuickLaTeX.com 2](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-19ca22a3e0d58f9f9adf5c72891640fa_l3.png)
![Rendered by QuickLaTeX.com a=1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-822be0dc3f5f78f900c375c8d41f55b3_l3.png)
![Rendered by QuickLaTeX.com b=2](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-0dbaead049d51e6d1ad39b30f499dbc5_l3.png)
![Rendered by QuickLaTeX.com y^2+3y+2=0 \to (y+1)(y+2)=0 \to \left\{ \begin{array}{l} y+1=0 \\ y+2=0 \end{array} \right. \to \left\{ \begin{array}{l} y=-1 \\ y=-2 \end{array} \right.](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-4e84cb2ced9920e67f7a870def5c1e56_l3.png)
Example 3:
Solve the quadratic equation ![Rendered by QuickLaTeX.com x^2-9=0.](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-dc4182544d009ba92ea598e73f5de10d_l3.png)
![Rendered by QuickLaTeX.com x^2-9=0.](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-dc4182544d009ba92ea598e73f5de10d_l3.png)
Assuming
and
, the identity
can be used to factor this equation. We have,
![Rendered by QuickLaTeX.com a=x](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-51fc686a349c3693edcf1669e7c72a9f_l3.png)
![Rendered by QuickLaTeX.com b=\sqrt{9}=3](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-893b9be723c0f46a951035836d8415b8_l3.png)
![Rendered by QuickLaTeX.com a^2 - b^2 = (a+b)(a-b)](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-cf3bd1117c363a2c103172b61adab7f8_l3.png)
![Rendered by QuickLaTeX.com x^2-9=0 \to x^2-3^2=0 \to (x+3)(x-3)=0\to \left\{ \begin{array}{l} x+3=0 \\ x-3=0 \end{array} \right. \to \left\{ \begin{array}{l} x=-3 \\ x=3 \end{array} \right.](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-aa3b5c960bf1105173520dfc0bf7034a_l3.png)
2) Guess one of the factors:
If coefficients are integer, it is sometimes easy to guess one of the factors.
Example 4: Solve the quadratic equation ![Rendered by QuickLaTeX.com 2x^2-3x+1=0](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-0ef535d6e062679a6f4496c02be4fe76_l3.png)
![Rendered by QuickLaTeX.com 2x^2-3x+1=0](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-0ef535d6e062679a6f4496c02be4fe76_l3.png)
Since
satisfies the equation,
is one of the factors. We can divide the equation by
to find the other factor:
![Rendered by QuickLaTeX.com x=1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-b56fab26e46772cf8ff34087818dffb0_l3.png)
![Rendered by QuickLaTeX.com x-1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-44a1b71d2bc7845e3cccd91bd17663e4_l3.png)
![Rendered by QuickLaTeX.com (x - 1)](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-51a84c61c967d0a66cbcc3e27559bb6c_l3.png)
![Rendered by QuickLaTeX.com \begin{array}{l | l} & x-1 \\ 2x^2-3x+1 &2x-1\\2x^2-2x & \\ -x+1&\\-x+1&\\ 0&\end{array}](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-2e365d82d647383b4307ab3c29d218f5_l3.png)
Therefore, the other factor is
and the original has two answers
and
.
![Rendered by QuickLaTeX.com 2x-1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-5e68d9daad903c514a6de64f2e92cecf_l3.png)
![Rendered by QuickLaTeX.com x=1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-b56fab26e46772cf8ff34087818dffb0_l3.png)
![Rendered by QuickLaTeX.com 2x-1=0 \to x= \frac{1}{2}](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-742e1efdf6f9ac5ea17d7e36efa44b25_l3.png)
Example 5: Solve the quadratic equation ![Rendered by QuickLaTeX.com -4x-3=x^2](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-6a748ad01f5bb538fcb5909926d8bf7f_l3.png)
![Rendered by QuickLaTeX.com -4x-3=x^2](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-6a748ad01f5bb538fcb5909926d8bf7f_l3.png)
![Rendered by QuickLaTeX.com -4x-3=x^2 \to x^2+4x+3=0](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-4d0e71d931af1f043cf291a3a4b06bc3_l3.png)
![Rendered by QuickLaTeX.com x=-1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-b95c31d03019c251ba4ada60f34091a0_l3.png)
![Rendered by QuickLaTeX.com x+1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-4fd3a5ddbe0b185c84b98f4d14b05374_l3.png)
![Rendered by QuickLaTeX.com (x+1)](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-f078d1a3bc244c876252165d3ea4443d_l3.png)
![Rendered by QuickLaTeX.com \begin{array}{l | l} & x+1 \\ x^2+4x+3 &x+3\\x^2+x & \\ 3x+3&\\3x+3&\\ 0 & \end{array}](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-d02bac3669d5bcdad99565d9c862f201_l3.png)
Thus, the other factor is
and
is the other answer. Therefore, the original has two answers
and
.
![Rendered by QuickLaTeX.com x+3](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-a3222430fd1381cc169eaf0d1dc2608c_l3.png)
![Rendered by QuickLaTeX.com x=-3](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-384331ae00e43d3652de08178fd59445_l3.png)
![Rendered by QuickLaTeX.com x=-1](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-b95c31d03019c251ba4ada60f34091a0_l3.png)
![Rendered by QuickLaTeX.com x=-3](https://www.solved-problems.com/wp-content/ql-cache/quicklatex.com-62307ad6f669b31b44e0b86092424f5b_l3.png)
As explained, factoring quadratic equations does not have a straight-forward procedure to follow. Therefore, the usage of this method is restricted to some trivial equations and relies mostly on your skills. You can learn about other methods of solving quadratic equations here.
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