2EXAMPLE 1Factor trinomials of the form x2 + bx + cFactor the expression.a. x2 – 9x + 20b. x2 + 3x – 12SOLUTIONa. You want x2 – 9x + 20 = (x + m)(x + n) where mn = 20 and m + n = –9.ANSWERNotice that m = –4 and n = –5.So, x2 – 9x + 20 = (x – 4)(x – 5).
3EXAMPLE 1Factor trinomials of the form x2 + bx + cb. You want x2 + 3x – 12 = (x + m)(x + n) where mn = – 12 and m + n = 3.ANSWERNotice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored.
4GUIDED PRACTICEfor Example 1Factor the expression. If the expression cannot be factored, say so.1. x2 – 3x – 182. n2 – 3n + 93. r2 + 2r – 63ANSWERANSWERANSWER(x – 6)(x + 3)cannot be factored(r + 9)(r –7)
5Factor with special patterns EXAMPLE 2Factor with special patternsFactor the expression.a. x2 – 49= x2 – 72Difference of two squares= (x + 7)(x – 7)
8Standardized Test Practice EXAMPLE 3Standardized Test PracticeSOLUTIONx2 – 5x – 36 = 0Write original equation.(x – 9)(x + 4) = 0Factor.x – 9 = 0 orx + 4 = 0Zero product propertyx = 9 orx = –4Solve for x.ANSWERThe correct answer is C.
9GUIDED PRACTICEfor Examples 3 and 48. Solve the equation x2 – x – 42 = 0.ANSWER–6 or 7
10Find the zeros of quadratic functions. EXAMPLE 5Find the zeros of quadratic functions.Find the zeros of the function by rewriting the function in intercept form.a. y = x2 – x – 12b. y = x2 + 12x + 36SOLUTIONa. y = x2 – x – 12Write original function.= (x + 3)(x – 4)Factor.The zeros of the function are –3 and 4.Check Graph y = x2 – x – 12. The graph passes through (–3, 0) and (4, 0).
11Find the zeros of quadratic functions. EXAMPLE 5Find the zeros of quadratic functions.b. y = x2 + 12x + 36Write original function.= (x + 6)(x + 6)Factor.The zeros of the function is –6Check Graph y = x2 + 12x The graph passes through ( –6, 0).
12GUIDED PRACTICEGUIDED PRACTICEfor Examplefor Example 5Find the zeros of the function by rewriting the function in intercept form.y = x2 + 5x – 14ANSWER–7 and 2y = x2 – 7x – 30ANSWER–3 and 10
14EXAMPLE 1Factor ax2 + bx + c where c > 0Factor 5x2 – 17x + 6.SOLUTIONYou want 5x2 – 17x + 6 = (kx + m)(lx + n) where k and l are factors of 5 and m and n are factors of 6. You can assume that k and l are positive and k ≥ l. Because mn > 0, m and n have the same sign. So, m and n must both be negative because the coefficient of x, –17, is negative.
15EXAMPLE 1Factor ax2 + bx + c where c > 0ANSWERThe correct factorization is 5x2 –17x + 6 = (5x – 2)(x – 3).
16EXAMPLE 2Factor ax2 + bx + c where c < 0Factor 3x2 + 20x – 7.SOLUTIONYou want 3x2 + 20x – 7 = (kx + m)(lx + n) where k and l are factors of 3 and m and n are factors of –7. Because mn < 0, m and n have opposite signs.ANSWERThe correct factorization is 3x2 + 20x – 7 = (3x – 1)(x + 7).
17GUIDED PRACTICEGUIDED PRACTICEfor Examples 1 and 2Factor the expression. If the expression cannot be factored, say so.x2 – 20x – 3ANSWER(7x + 1)(x – 3)w2 + w + 3ANSWERcannot be factored