# Factor a Quadratic Equation by Grouping

Any quadratic equation with rational roots can be factored by the grouping method. Here are the steps, followed by a few examples.

**Steps:**

- Use algebra to put the equation in standard form with all terms on one side and 0 on the other.
- Divide both sides by the greatest common factor (GCF), if any, to get to lowest terms.
- Multiply the A
coefficient (the squared term) by the C coefficient (the constant), to find AC. Note: "A" and "C" come from Ax
^{2}+Bx+C=0, which is the standard form of a quadratic equation. - Find the two integers that multiply together to equal AC and also add together to equal B (the coefficient of the x term).
- Rewrite the B term as two new terms using the pair of numbers just found.
- Group so that one x term falls into each resultant binomial.
- Factor each pair of terms.
- Factor again -- you should now have a fully factored expression.
- Set each factor equal to zero and solve.
- Check your answers.

**Example: 4x ^{2}-13x=x+30**

- Subtract x+30 from both sides to get 4x
^{2}-14x - 30 = 0. - The GCF is 2, so divide each term by 2. The equation is now
2x
^{2}-7x-15=0. - 2 * (-15) = -30, so AC is -30.
- The integer pairs that multiply together to equal -30 are: (-1, 30), (-2, 15), (-3,
10), (-5, 6), (1, -30), (-2, 15), (3, -10), and (5, -6)

We want the pair that adds together to equal -7. Since 3 + (-10) = -7, we choose 3 and -10 as our two numbers. - Rewrite
the middle term of our equation using the two numbers just found. The
middle term was -7x, so substitute 3x - 10x for the -7x, giving the
equation as 2x
^{2}+ 3x - 10x - 15=0. - Group to get (2x
^{2}+3x)+(-10x-15)=0 - Factor each pair of terms: x(2x+3) - 5(2x+3)=0
- Factor out the common factor (2x + 3) to get (2x+3)(x-5)=0
- Solve each factor

When 2x + 3 = 0, x = -3/2.

When x - 5 = 0, x = 5.

Therefore, x is either -3/2 or 5 - Check using the original equation:

4(-3/2)^{2}- 13(-3/2) = (-3/2) + 30, so 9 + 39/2 = 57/2, and 57/2 = 57/2. Therefore x = -3/2 is a solution.

4(5)^{2}- 13(5) = (5) + 30, so 100 - 65 = 35, and 35 = 35. Therefore x = 5 is a solution and the quadratic has been solved.

**Example: x ^{2}- 2 = 0**

- This is already in standard form, although you could write it x
^{2}+ 0x - 2 = 0. - The GCF is 1, so no need to divide.
- AC = -2
- Possible integer pairs are (-1, 2) and (1, -2). Neither pair sums to 0 (our B coefficient), so we now know that x
^{2}- 2 = 0 does not have rational roots.

Result: x^{2} - 2 = 0 is not factorable in the rational number system. (It can be factored in the reals.)

**Example: 6y ^{2} + 11y - 35 = 0**

- Equation is already in standard form.
- Terms are relatively prime (GCF = 1), so no need to divide out common factors.
- AC = -210
- Possible pairs: (-1, 210), (-2, 105), (-3, 70), (-5, 22), (-6, 35), (-7, 30), (-10, 21) <<< stop here because -10 + 21 = 11, which is what we want!
- 6y
^{2}- 10y + 21y - 35 = 0 - (6y
^{2}- 10y) + (21y - 35) = 0 - 2y(3y - 5) + 7(3y - 5) = 0
- (3y - 5)(2y + 7) = 0
- 3y - 5 = 0 implies that y = 5/3

2y + 7 = 0 implies that y = -7/2

Therefore y is 5/3 or -7/2. - Does 6(5/3)
^{2}+ 11(5/3) - 35 = 0? Yes, true.

Does 6(-7/2)^{2}+ 11(-7/2) - 35 = 0? Yes, true.

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