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# EXAMPLE 2

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1. EXAMPLE 2 Find a least common multiple (LCM) Find the least common multiple of 4x2 –16 and 6x2 –24x + 24. SOLUTION STEP 1 Factor each polynomial. Write numerical factors asproducts of primes. 4x2 – 16 = 4(x2 – 4) = (22)(x + 2)(x – 2) 6x2 – 24x + 24 = 6(x2 – 4x + 4) = (2)(3)(x – 2)2

2. EXAMPLE 2 Find a least common multiple (LCM) STEP 2 Form the LCM by writing each factor to the highest power it occurs in either polynomial. LCM = (22)(3)(x + 2)(x – 2)2 = 12(x + 2)(x – 2)2

3. x + 1 7 x x + 1 + 9x2 3x2 + 3x 7 7 x x = + + 9x2 9x2 3x2 + 3x 3x(x + 1) 7 3x x 9x2 3x + 3x(x + 1) EXAMPLE 3 Add with unlike denominators Add: SOLUTION To find the LCD, factor each denominator and write each factor to the highest power it occurs. Note that9x2 = 32x2and3x2 + 3x = 3x(x + 1), so the LCD is 32x2 (x + 1) = 9x2(x 1 1). Factor second denominator. LCD is 9x2(x + 1).

4. 3x2 7x + 7 = + 9x2(x + 1) 9x2(x + 1) 3x2+ 7x + 7 = 9x2(x + 1) EXAMPLE 3 Add with unlike denominators Multiply. Add numerators.

5. x – 3 x + 2 x + 2 –2x –1 –2x –1 – x– 3 2x – 2 2x – 2 x2 – 4x + 3 x2 – 4x + 3 – x + 2 x + 2 – 2x – 1 – 2x – 1 – = (x – 1)(x – 3) (x – 1)(x – 3) 2(x – 1) 2(x – 1) – = 2 x2 – x – 6 – 4x – 2 – 2 = 2(x – 1)(x – 3) 2(x – 1)(x – 3) EXAMPLE 4 Subtract with unlike denominators Subtract: SOLUTION Factor denominators. LCD is 2(x  1)(x  3). Multiply.

6. x2 – x – 6 – (– 4x – 2) = 2(x – 1)(x – 3) x2+ 3x – 4 = 2(x – 1)(x – 3) (x –1)(x + 4) = 2(x – 1)(x – 3) x + 4 = 2(x –3) EXAMPLE 4 Subtract with unlike denominators Subtract numerators. Simplify numerator. Factor numerator. Divide out common factor. Simplify.

7. STEP 1 Factor each polynomial. Write numerical factors asproducts of primes. STEP 2 Form the LCM by writing each factor to the highest power it occurs in either polynomial. for Examples 2, 3 and 4 GUIDED PRACTICE Find the least common multiple of the polynomials. 5. 5x3and 10x2–15x 5x3 = 5(x) (x2) 10x2 – 15x= 5(x) (2x – 3) LCM = 5x3 (2x – 3)

8. STEP 1 Factor each polynomial. Write numerical factors asproducts of primes. 23(x – 2) 8x – 16 = 8(x – 2) = 4 3(x – 2 )(x + 3) 12x2 + 12x – 72 = 12(x2 + x – 6) = STEP 2 Form the LCM by writing each factor to the highest power it occurs in either polynomial. LCM = 8 3(x – 2)(x + 3) for Examples 2, 3 and 4 GUIDED PRACTICE Find the least common multiple of the polynomials. 6. 8x – 16 and 12x2 + 12x – 72 = 24(x – 2)(x + 3)

9. 3 3 1 1 7. – – 4x 4x 7 7 3 1 4x 7 – 7 4x 4x 7 – 4x 21 7(4x) 4x(7) = 21 – 4x 28x for Examples 2, 3 and 4 GUIDED PRACTICE SOLUTION LCDis 28x Multiply Simplify

10. x x 1 1 8. + + 3x2 3x2 9x2 – 12x 9x2 – 12x 1 = + x 3x2 3x(3x – 4) 1 = + x 3x – 4 3x2 3x(3x – 4) 3x – 4 3x – 4 x + x2 x 3x2 (3x – 4) = 3x2 (3x – 4) for Examples 2, 3 and 4 GUIDED PRACTICE SOLUTION Factor denominators LCD is 3x2 (3x – 4) Multiply

11. 3x – 4 + x2 3x2 (3x – 4) x2 + 3x – 4 3x2 (3x – 4) for Examples 2, 3 and 4 GUIDED PRACTICE Add numerators Simplify

12. x x 5 5 9. + + x2 – x – 12 x2 – x – 12 12x – 48 12x – 48 x = + 5 (x+3)(x – 4) 12 (x – 4) x + 3 5 = + 12 x + 3 x 12(x – 4) (x + 3)(x – 4) 12 5(x + 3) 12x = + 12(x + 3)(x – 4) 12(x + 3)(x – 4) for Examples 2, 3 and 4 GUIDED PRACTICE SOLUTION Factor denominators LCD is 12(x – 4) (x+3) Multiply

13. 12x + 5x + 15 = 12(x + 3)(x – 4) = 17x + 15 12(x +3)(x + 4) for Examples 2, 3 and 4 GUIDED PRACTICE Add numerators Simplify

14. x + 1 x + 1 6 6 10. – – 6 x + 1 – x2 + 4x + 4 x2 + 4x + 4 x2 – 4 x2 – 4 = (x – 2)(x + 2) x + 2 x – 2 (x+ 2)(x + 2) x + 2 x – 2 6 – = (x – 2)(x + 2) x + 1 (x + 2)(x + 2) 6x + 12 x2 – 2x + x – 2 (x – 2)(x + 2)(x + 2) – = (x + 2)(x + 2)(x – 2) for Examples 2, 3 and 4 GUIDED PRACTICE SOLUTION Factor denominators LCD is (x – 2) (x+2)2 Multiply

15. = x2 – 2x + x – 2 – (6x + 12) = (x + 2)2(x – 4) x2 – 7x –14 (x + 2)2 (x – 2) for Example 2, 3 and 4 GUIDED PRACTICE Subtract numerators Simplify