19/02/2014 CH.7.3 Factoring x2 + bx + c

Warm Up 1. Which pair of factors of 8 has a sum of 9? 2. Which pair of factors of 30 has a sum of –17? Multiply. 1 and 8 –2 and –15 3. (x +2)(x +3) x2 + 5x + 6 4. (r + 5)(r – 9) r2 – 4r – 45

Objective Factor quadratic trinomials of the form x2 + bx + c.

In Chapter 7, you learned how to multiply two binomials using the Distributive Property or the FOIL method. In this lesson, you will learn how to factor a trinomial into two binominals.

Notice that when you multiply (x + 2)(x + 5), the constant term in the trinomial is the product of the constants in the binomials. (x + 2)(x + 5) = x2 + 7x + 10 You can use this fact to factor a trinomial into its binomial factors. Look for two numbers that are factors of the constant term in the trinomial. Write two binomials with those numbers, and then multiply to see if you are correct.

Example 1A: Factoring Trinomials by Guess and Check Factor x2 + 15x + 36 by guess and check. ( + )( + ) Write two sets of parentheses. (x + )(x + ) The first term is x2, so the variable terms have a coefficient of 1. The constant term in the trinomial is 36.  Try factors of 36 for the constant terms in the binomials. (x + 1)(x + 36) = x2 + 37x + 36  (x + 2)(x + 18) = x2 + 20x + 36  (x + 3)(x + 12) = x2 + 15x + 36 The factors of x2 + 15x + 36 are (x + 3)(x + 12). x2 + 15x + 36 = (x + 3)(x + 12)

When you multiply two binomials, multiply: First terms Outer terms Inner terms Last terms Remember!

    Check It Out! Example 1a Factor each trinomial by guess and check. x2 + 10x + 24 ( + )( + ) Write two sets of parentheses. (x + )(x + ) The first term is x2, so the variable terms have a coefficient of 1. The constant term in the trinomial is 24.  (x + 1)(x + 24) = x2 + 25x + 24 Try factors of 24 for the constant terms in the binomials.  (x + 2)(x + 12) = x2 + 14x + 24  (x + 3)(x + 8) = x2 + 11x + 24  (x + 4)(x + 6) = x2 + 10x + 24 The factors of x2 + 10x + 24 are (x + 4)(x + 6). x2 + 10x + 24 = (x + 4)(x + 6)

   Check It Out! Example 1b Factor each trinomial by guess and check. x2 + 7x + 12 ( + )( + ) Write two sets of parentheses. (x + )(x + ) The first term is x2, so the variable terms have a coefficient of 1. The constant term in the trinomial is 12.  (x + 1)(x + 12) = x2 + 13x + 12 Try factors of 12 for the constant terms in the binomials.  (x + 2)(x + 6) = x2 + 8x + 12  (x + 3)(x + 4) = x2 + 7x + 12 The factors of x2 + 7x + 12 are (x + 3)(x + 4). x2 + 7x + 12 = (x + 3)(x + 4)

The guess and check method is usually not the most efficient method of factoring a trinomial. Look at the product of (x + 3) and (x + 4). x2 12 (x + 3)(x +4) = x2 + 7x + 12 3x 4x The coefficient of the middle term is the sum of 3 and 4. The third term is the product of 3 and 4.

When c is positive, its factors have the same sign When c is positive, its factors have the same sign. The sign of b tells you whether the factors are positive or negative. When b is positive, the factors are positive and when b is negative, the factors are negative.

Example 2A: Factoring x2 + bx + c When c is Positive Factor each trinomial. Check your answer. x2 + 6x + 5 (x + )(x + ) b = 6 and c = 5; look for factors of 5 whose sum is 6. Factors of 5 Sum 1 and 5 6  The factors needed are 1 and 5. (x + 1)(x + 5) Check (x + 1)(x + 5) = x2 + x + 5x + 5 Use the FOIL method. = x2 + 6x + 5  The product is the original polynomial.

Example 2B: Factoring x2 + bx + c When c is Positive Factor each trinomial. Check your answer. x2 + 6x + 9 (x + )(x + ) b = 6 and c = 9; look for factors of 9 whose sum is 6. Factors of 9 Sum 1 and 9 10  3 and 3 6  The factors needed are 3 and 3. (x + 3)(x + 3) Check (x + 3)(x + 3 ) = x2 +3x + 3x + 9 Use the FOIL method. = x2 + 6x + 9  The product is the original polynomial.

Example 2C: Factoring x2 + bx + c When c is Positive Factor each trinomial. Check your answer. x2 – 8x + 15 (x + )(x + ) b = –8 and c = 15; look for factors of 15 whose sum is –8. Factors of 15 Sum –1 and –15 –16  –3 and –5 –8  The factors needed are –3 and –5 . (x – 3)(x – 5) Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15 Use the FOIL method. = x2 – 8x + 15  The product is the original polynomial.

   Check It Out! Example 2a Factor each trinomial. Check your answer. x2 + 8x + 12 (x + )(x + ) b = 8 and c = 12; look for factors of 12 whose sum is 8. Factors of 12 Sum 1 and 12 13  2 and 6 8  The factors needed are 2 and 6 . (x + 2)(x + 6) Check (x + 2)(x + 6 ) = x2 + 2x + 6x + 12 Use the FOIL method. = x2 + 8x + 12  The product is the original polynomial.

   Check It Out! Example 2b Factor each trinomial. Check your answer. x2 – 5x + 6 (x + )(x+ ) b = –5 and c = 6; look for factors of 6 whose sum is –5. Factors of 6 Sum –1 and –6 –7  –2 and –3 –5  The factors needed are –2 and –3. (x – 2)(x – 3) Check (x – 2)(x – 3) = x2 –2x – 3x + 6 Use the FOIL method. = x2 – 5x + 6  The product is the original polynomial.

   Check It Out! Example 2c Factor each trinomial. Check your answer. x2 + 13x + 42 (x + )(x + ) b = 13 and c = 42; look for factors of 42 whose sum is 13. Factors of 42 Sum 1 and 42 43  6 and 7 13  2 and 21 23 The factors needed are 6 and 7. (x + 6)(x + 7) Check (x + 6)(x + 7) = x2 + 7x + 6x + 42 Use the FOIL method. = x2 + 13x + 42  The product is the original polynomial.

   Check It Out! Example 2d Factor each trinomial. Check your answer. x2 – 13x + 40 b = –13 and c = 40; look for factors of 40 whose sum is –13. (x + )(x+ ) Factors of 40 Sum  –2 and –20 –22 –4 and –10 –14  –5 and –8 –13 The factors needed are –5 and –8. (x – 5)(x – 8) Check (x – 5)(x – 8) = x2 – 5x – 8x + 40 Use the FOIL method. = x2 – 13x + 40  The product is the original polynomial.

When c is negative, its factors have opposite signs When c is negative, its factors have opposite signs. The sign of b tells you which factor is positive and which is negative. The factor with the greater absolute value has the same sign as b.

Example 3A: Factoring x2 + bx + c When c is Negative Factor each trinomial. x2 + x – 20 b = 1 and c = –20; look for factors of –20 whose sum is 1. The factor with the greater absolute value is positive. (x + )(x + ) Factors of –20 Sum  –1 and 20 19 –2 and 10 8  –4 and 5 1 The factors needed are +5 and –4. (x – 4)(x + 5)

Example 3B: Factoring x2 + bx + c When c is Negative Factor each trinomial. x2 – 3x – 18 b = –3 and c = –18; look for factors of –18 whose sum is –3. The factor with the greater absolute value is negative. (x + )(x + ) Factors of –18 Sum  1 and –18 –17 2 and – 9 – 7  3 and – 6 – 3 The factors needed are 3 and –6. (x – 6)(x + 3)

If you have trouble remembering the rules for which factor is positive and which is negative, you can try all the factor pairs and check their sums. Helpful Hint

  Check It Out! Example 3a Factor each trinomial. Check your answer. x2 + 2x – 15 b = 2 and c = –15; look for factors of –15 whose sum is 2. The factor with the greater absolute value is positive. (x + )(x + ) Factors of –15 Sum  –1 and 15 14 –3 and 5 2  The factors needed are –3 and 5. (x – 3)(x + 5)

  Check It Out! Example 3b Factor each trinomial. Check your answer. x2 – 6x + 8 (x + )(x + ) b = –6 and c = 8; look for factors of 8 whose sum is –6. Factors of 8 Sum  –1 and –6 –7 –2 and –4 –6  The factors needed are –4 and –2. (x – 2)(x – 4)

  Check It Out! Example 3c Factor each trinomial. Check your answer. X2 – 8x – 20 (x + )(x + ) b = –8 and c = –20; look for factors of –20 whose sum is –8. The factor with the greater absolute value is negative. Factors of –20 Sum  1 and –20 –19 2 and –10 –8  The factors needed are –10 and 2. (x – 10)(x + 2)

A polynomial and the factored form of the polynomial are equivalent expressions. When you evaluate these two expressions for the same value of the variable, the results are the same.

Example 4A: Evaluating Polynomials Factor y2 + 10y + 21. Show that the original polynomial and the factored form have the same value for n = 0, 1, 2, 3, and 4. y2 + 10y + 21 (y + )(y + ) b = 10 and c = 21; look for factors of 21 whose sum is 10. Factors of 21 Sum 1 and 21 21  3 and 7 10  The factors needed are 3 and 7. (y + 3)(y + 7)

Example 4A Continued Evaluate the original polynomial and the factored form for n = 0, 1, 2, 3, and 4. (y + 7)(y + 3) (0 + 7)(0 + 3) = 21 (1 + 7)(1 + 3) = 32 (2 + 7)(2 + 3) = 45 (3 + 7)(3 + 3) = 60 (4 + 7)(4 + 3) = 77 y 1 2 3 4 y2 + 10y + 21 02 + 10(0) + 21 = 21 y 1 2 3 4 12 + 10(1) + 21 = 32 22 + 10(2) + 21 = 45 32 + 10(3) + 21 = 60 42 + 10(4) + 21 = 77 The original polynomial and the factored form have the same value for the given values of n.

Check It Out! Example 4 Factor n2 – 7n + 10. Show that the original polynomial and the factored form have the same value for n = 0, 1, 2, 3, and 4. n2 – 7n + 10 (n + )(n + ) b = –7 and c = 10; look for factors of 10 whose sum is –7. Factors of 10 Sum –1 and –10 –11  –2 and –5 – 7  The factors needed are –2 and –5. (n – 5)(n – 2)

Check It Out! Example 4 Continued Evaluate the original polynomial and the factored form for n = 0, 1, 2, 3, and 4. (n – 5)(n – 2 ) (0 – 5)(0 – 2) = 10 n 1 2 3 4 (1 – 5)(1 – 2) = 4 (2 – 5)(2 – 2) = 0 (3 – 5)(3 – 2) = –2 (4 – 5)(4 – 2) = –2 n2 –7n + 10 02 –7(0) + 10 = 10 y 1 2 3 4 32 – 7(3) + 10 = –2 22 – 7(2) + 10 = 0 42 – 7(4) + 10 = –2 12 – 7(1) + 10 = 4 The original polynomial and the factored form have the same value for the given values of n.

Lesson Quiz: Part I Factor each trinomial. 1. x2 – 11x + 30 2. x2 + 10x + 9 3. x2 – 6x – 27 4. x2 + 14x – 32 (x – 5)(x – 6) (x + 1)(x + 9) (x – 9)(x + 3) (x + 16)(x – 2)

Lesson Quiz: Part II Factor n2 + n – 6. Show that the original polynomial and the factored form have the same value for n = 0, 1, 2, 3 ,and 4. (n + 3)(n – 2)

Warm Up Solve for x. 1. 16x – 3 = 12x + 13 2. 2x – 4 = 90 ABCD is a parallelogram. Find each measure. 3. CD 4. mC 4 47 14 104°

Warm Up 5 –1 1. Find AB for A (–3, 5) and B (1, 2). 2. Find the slope of JK for J(–4, 4) and K(3, –3). ABCD is a parallelogram. Justify each statement. 3. ABC  CDA 4. AEB  CED 5 –1  opp. s  Vert. s Thm.

Objective Prove that a given quadrilateral is a rectangle, rhombus, or square.

When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle.

Example 1: Carpentry Application A manufacture builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle? Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.

Check It Out! Example 1 A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle? Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one  of WXYZ is a right . If one angle is a right , then by Theorem 6-5-1 the frame is a rectangle.

Below are some conditions you can use to determine whether a parallelogram is a rhombus.

In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram. Caution To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43.

You can also prove that a given quadrilateral is a rectangle, rhombus, or square by using the definitions of the special quadrilaterals. Remember!

Example 2A: Applying Conditions for Special Parallelograms Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a rhombus. The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

Example 2B: Applying Conditions for Special Parallelograms Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a square. Step 1 Determine if EFGH is a parallelogram. Given Quad. with diags. bisecting each other  EFGH is a parallelogram.

Example 2B Continued Step 2 Determine if EFGH is a rectangle. Given. EFGH is a rectangle. with diags.   rect. Step 3 Determine if EFGH is a rhombus. with one pair of cons. sides   rhombus EFGH is a rhombus.

Example 2B Continued Step 4 Determine is EFGH is a square. Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition. The conclusion is valid.

Check It Out! Example 2 Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: ABC is a right angle. Conclusion: ABCD is a rectangle. The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .

Example 3A: Identifying Special Parallelograms in the Coordinate Plane Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)

Example 3A Continued Step 1 Graph PQRS.

Example 3A Continued Step 2 Find PR and QS to determine if PQRS is a rectangle. Since , the diagonals are congruent. PQRS is a rectangle.

Example 3A Continued Step 3 Determine if PQRS is a rhombus. Since , PQRS is a rhombus. Step 4 Determine if PQRS is a square. Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.

Example 3B: Identifying Special Parallelograms in the Coordinate Plane Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3) Step 1 Graph WXYZ.

Example 3B Continued Step 2 Find WY and XZ to determine if WXYZ is a rectangle. Since , WXYZ is not a rectangle. Thus WXYZ is not a square.

Example 3B Continued Step 3 Determine if WXYZ is a rhombus. Since (–1)(1) = –1, , WXYZ is a rhombus.

Check It Out! Example 3A Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)

Check It Out! Example 3A Continued Step 1 Graph KLMN.

Check It Out! Example 3A Continued Step 2 Find KM and LN to determine if KLMN is a rectangle. Since , KMLN is a rectangle.

Check It Out! Example 3A Continued Step 3 Determine if KLMN is a rhombus. Since the product of the slopes is –1, the two lines are perpendicular. KLMN is a rhombus.

Check It Out! Example 3A Continued Step 4 Determine if KLMN is a square. Since KLMN is a rectangle and a rhombus, it has four right angles and four congruent sides. So KLMN is a square by definition.

Check It Out! Example 3B Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)

Check It Out! Example 3B Continued Step 1 Graph PQRS.

Check It Out! Example 3B Continued Step 2 Find PR and QS to determine if PQRS is a rectangle. Since , PQRS is not a rectangle. Thus PQRS is not a square.

Check It Out! Example 3B Continued Step 3 Determine if PQRS is a rhombus. Since (–1)(1) = –1, are perpendicular and congruent. PQRS is a rhombus.

Lesson Quiz: Part I 1. Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

Lesson Quiz: Part II 2. Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: PQRS and PQNM are parallelograms. Conclusion: MNRS is a rhombus. valid

Lesson Quiz: Part III 3. Use the diagonals to determine whether a parallelogram with vertices A(2, 7), B(7, 9), C(5, 4), and D(0, 2) is a rectangle, rhombus, or square. Give all the names that apply. AC ≠ BD, so ABCD is not a rect. or a square. The slope of AC = –1, and the slope of BD = 1, so AC  BD. ABCD is a rhombus.

Objectives Prove and apply properties of rectangles, rhombuses, and squares. Use properties of rectangles, rhombuses, and squares to solve problems.

Vocabulary rectangle rhombus square

A second type of special quadrilateral is a rectangle A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

Example 1: Craft Application A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM. Rect.  diags.  KM = JL = 86 Def. of  segs.  diags. bisect each other Substitute and simplify.

Check It Out! Example 1a Carpentry The rectangular gate has diagonal braces. Find HJ. Rect.  diags.  HJ = GK = 48 Def. of  segs.

Check It Out! Example 1b Carpentry The rectangular gate has diagonal braces. Find HK. Rect.  diags.  Rect.  diagonals bisect each other JL = LG Def. of  segs. JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

A rhombus is another special quadrilateral A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

Like a rectangle, a rhombus is a parallelogram Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

Example 2A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. WV = XT Def. of rhombus 13b – 9 = 3b + 4 Substitute given values. 10b = 13 Subtract 3b from both sides and add 9 to both sides. b = 1.3 Divide both sides by 10.

Example 2A Continued TV = XT Def. of rhombus Substitute 3b + 4 for XT. TV = 3b + 4 TV = 3(1.3) + 4 = 7.9 Substitute 1.3 for b and simplify.

Example 2B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find mVTZ. mVZT = 90° Rhombus  diag.  14a + 20 = 90° Substitute 14a + 20 for mVTZ. Subtract 20 from both sides and divide both sides by 14. a = 5

Example 2B Continued Rhombus  each diag. bisects opp. s mVTZ = mZTX mVTZ = (5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ = [5(5) – 5)]° = 20° Substitute 5 for a and simplify.

Check It Out! Example 2a CDFG is a rhombus. Find CD. CG = GF Def. of rhombus 5a = 3a + 17 Substitute a = 8.5 Simplify GF = 3a + 17 = 42.5 Substitute CD = GF Def. of rhombus CD = 42.5 Substitute

Check It Out! Example 2b CDFG is a rhombus. Find the measure. mGCH if mGCD = (b + 3)° and mCDF = (6b – 40)° mGCD + mCDF = 180° Def. of rhombus b + 3 + 6b – 40 = 180° Substitute. 7b = 217° Simplify. b = 31° Divide both sides by 7.

Check It Out! Example 2b Continued mGCH + mHCD = mGCD Rhombus  each diag. bisects opp. s 2mGCH = mGCD 2mGCH = (b + 3) Substitute. 2mGCH = (31 + 3) Substitute. mGCH = 17° Simplify and divide both sides by 2.

A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms. Helpful Hint

Example 3: Verifying Properties of Squares Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

Example 3 Continued Step 1 Show that EG and FH are congruent. Since EG = FH,

Example 3 Continued Step 2 Show that EG and FH are perpendicular. Since ,

Example 3 Continued Step 3 Show that EG and FH are bisect each other. Since EG and FH have the same midpoint, they bisect each other. The diagonals are congruent perpendicular bisectors of each other.

SV ^ TW SV = TW = 122 so, SV @ TW . slope of TW = –11 1 slope of SV = Check It Out! Example 3 The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3) , and W(1, –9) . Show that the diagonals of square STVW are congruent perpendicular bisectors of each other. SV = TW = 122 so, SV @ TW . 1 11 slope of SV = slope of TW = –11 SV ^ TW

Check It Out! Example 3 Continued Step 1 Show that SV and TW are congruent. Since SV = TW,

Check It Out! Example 3 Continued Step 2 Show that SV and TW are perpendicular. Since

Check It Out! Example 3 Continued Step 3 Show that SV and TW bisect each other. Since SV and TW have the same midpoint, they bisect each other. The diagonals are congruent perpendicular bisectors of each other.

Example 4: Using Properties of Special Parallelograms in Proofs Given: ABCD is a rhombus. E is the midpoint of , and F is the midpoint of . Prove: AEFD is a parallelogram.

Example 4 Continued ||

Check It Out! Example 4 Given: PQTS is a rhombus with diagonal Prove:

Check It Out! Example 4 Continued Statements Reasons 1. PQTS is a rhombus. 1. Given. 2. Rhombus → each diag. bisects opp. s 2. 3. QPR  SPR 3. Def. of  bisector. 4. 4. Def. of rhombus. 5. 5. Reflex. Prop. of  6. 6. SAS 7. 7. CPCTC

Lesson Quiz: Part I A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length. 1. TR 2. CE 35 ft 29 ft

Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3. QP 4. mQRP 42 51°

Lesson Quiz: Part III 5. The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

Lesson Quiz: Part IV 6. Given: ABCD is a rhombus. Prove: 