**1.** |
To factor *x*^{2} + 12*x* + 35, begin by listing the positive and negative factors of the first and last terms: |

*x*^{2}: –*x*, *x* |

35: –35, –7, –5, –1, 1, 5, 7, 35 |

*x*^{2} is the square of either *x* or –*x*. Begin by trying *x* as the first term in each binomial: |

(*x* + __)(*x* + __) |

The coefficients of each *x* term are 1. The last terms of each binomial must multiply to 35 and add to 12, since 12*x* is the sum of the products of the Outside and Inside terms. 12*x* and 35 are positive, so we are looking for two positive numbers that multiply to 35 and add to 12. |

(1)(35) = 35, but 1 + 35 = 36 |

(5)(7) = 35, and 5 + 7 = 12 |

The constant of one binomial is 5 and the constant of the other binomial is 7: |

(*x* + 5)(*x* + 7) |

Check the answer using FOIL: |

First: (*x*)(*x*) = *x*^{2} |

Outside: (*x*)(7) = 7*x* |

Inside: (5)(*x*) = 5*x* |

Last: (5)(7) = 35 |

*x*^{2} + 7*x* + 5*x* + 35 = *x*^{2} + 12*x* + 35 |

**2.** |
To factor *x*^{2} – 6*x* + 8, begin by listing the positive and negative factors of the first and last terms: |

*x*^{2}: –*x*, *x* |

8: –8, –4, –2, –1, 1, 2, 4, 8 |

*x*^{2} is the square of either *x* or –*x*. Begin by trying *x* as the first term in each binomial: |

(*x* + __)(*x* + __) |

The coefficients of each *x* term are 1. The last terms of each binomial must multiply to 8 and add to –6, since –6*x* is the sum of the products of the Outside and Inside terms. –6*x* is negative and 8 is positive, so we are looking for two negative numbers that multiply to 8 and add to –6. |

(–1)(–8) = 8, but –1 + (–8) = –9 |

(–2)(–4) = 8, and –2 + (–4) = –6 |

The constant of one binomial is –2 and the constant of the other binomial is –4: |

(*x* – 2)(*x* – 4) |

Check the answer using FOIL: |

First: (*x*)(*x*) = *x*^{2} |

Outside: (*x*)(–4) = –4*x* |

Inside: (–2)(*x*) = –2*x* |

Last: (–2)(–4) = 8 |

*x*^{2} – 4*x* – 2*x* + 8 = *x*^{2} – 6*x* + 8 |

**3.** |
To factor *x*^{2} – *x* – 12, begin by listing the positive and negative factors of the first and last terms: |

*x*^{2}: –*x*, *x* |

12: –12, –6, –4, –3, –2, –1, 1, 2, 3, 4, 6, 12 |

*x*^{2} is the square of either *x* or –*x*. Begin by trying *x* as the first term in each binomial: |

(*x* + __)(*x* + __) |

The coefficients of each *x* term are 1. The last terms of each binomial must multiply to –12 and add to –1, because –*x* is the sum of the products of the Outside and Inside terms. –*x* is negative, as is –12, so we are looking for one positive number and one negative number that multiply to –12 and add to –1. |

(–12)(1) = –12, but –12 + 1 = –11 |

(12)(–1) = –12, but 12 + (–1) = 11 |

(–6)(2) = –12, but –6 + 2 = –4 |

(6)(–2) = –12, but 6 + (–2) = 4 |

(4)(–3) = –12, but 4 + (–3) = 1 |

(–4)(3) = –12, and –4 + 3 = –1 |

The constant of one binomial is –4 and the constant of the other binomial is 3: |

(*x* – 4)(*x* + 3) |

Check the answer using FOIL: |

First: (*x*)(*x*) = *x*^{2} |

Outside: (*x*)(3) = 3*x* |

Inside: (–4)(*x*) = –4*x* |

Last: (–4)(3) = –12 |

*x*^{2} + 3*x* – 4*x* – 12 = *x*^{2} – *x* – 12 |

**4.** |
To factor 2*x*^{2} + 3*x* + 1, begin by listing the positive and negative factors of the first and last terms: |

2*x*^{2}: –2, –1, 1, 2, –*x*, *x* |

1: –1, 1 |

2*x*^{2} could be (2*x*)(*x*) or (–2*x*)(–*x*). Begin by trying 2*x* as the first term in one binomial and *x* as the first term in the other binomial: |

(2*x* + __)(*x* + __) |

The coefficient of one *x* term is 2 and the coefficient of the other *x* term is 1. The last terms of each binomial must multiply to 1. One of those terms will be multiplied by 2*x*, and the other will be multiplied by 1*x*. The sum of those products will add to 3*x*. 3*x* and 1 are positive, so we are looking for two positive numbers that multiply to 1. The only possibility is 1 times 1, so the constant of each binomial is 1: |

(2*x* + 1)(*x* + 1) |

Check the answer using FOIL: |

First: (2*x*)(*x*) = 2*x*^{2} |

Outside: (2*x*)(1) = 2*x* |

Inside: (1)(*x*) = *x* |

Last: (1)(1) = 1 |

2*x*^{2} + 2*x* + *x* + 1 = 2*x*^{2} + 3*x* + 1 |

**5.** |
To factor 3*x*^{2} + 8*x* – 11, begin by listing the positive and negative factors of the first and last terms: |

3*x*^{2}: –3, –1, 1, 3, –*x*, *x* |

11: –11, –1, 1, 11 |

3*x*^{2} could be (3*x*)(*x*) or (–3*x*)(–*x*). Begin by trying 3*x* as the first term in one binomial and *x* as the first term in the other binomial: |

(3*x* + __)(*x* + __) |

The coefficient of one *x* term is 3, and the coefficient of the other *x* term is 1. The last terms of each binomial must multiply to –11. One of those terms will be multiplied by 3*x* and the other will be multiplied by 1*x*. The sum of those products will add to 8*x*. 8*x* is positive, but –11 is negative, so we are looking for a positive number and a negative number that multiply to –11. The possibilities are (–1)(11) and (1)(–11): |

(3*x* – 11)(*x* + 1) |

Check the answer using FOIL: |

First: (3*x*)(*x*) = 3*x*^{2} |

Outside: (3*x*)(1) = 3*x* |

Inside: (–11)(*x*) = –11*x* |

Last: (–11)(1) = –11 |

3*x*^{2} + 3*x* – 11*x* – 11 = 3*x*^{2} – 8*x* – 11, not 3*x*^{2} + 8*x* – 11 |

Try changing the signs of the last terms of each binomial: |

(3*x* + 11)(*x* – 1) |

Check the answer using FOIL: |

First: (3*x*)(*x*) = 3*x*^{2} |

Outside: (3*x*)(–1) = –3*x* |

Inside: (11)(*x*) = 11*x* |

Last: (11)(–1) = –11 |

3*x*^{2} – 3*x* + 11*x* – 11 = 3*x*^{2} + 8*x* – 11 |

Therefore, 3*x*^{2} + 8*x* – 11 factors to (3*x* + 11)(*x* – 1). |