8类RD Sharma解决方案–第7章分解–练习7.9

通过使用完成平方的方法，分解以下每个二次多项式

问题1. p ² + 6p + 8

解决方案：

p² + 6p + 8

= p² + 2 x p x 3 + 3² – 3² + 8   (completing the square)

= (p² + 6p + 3²) – 1

= (p + 3)² – 1

= (p + 3)² – (1)²       { ∵ a² + b²  = (a + b) (a – b)}

= (p + 3 + 1) (p + 3 – 1)

= (p + 4) (p + 2)

问题2 Q ² – 10Q + 21

解决方案：

q² – 10q + 21

= (q)² – 2 x q x 5 + (5)² – (5)² + 21   (completing the square)

= (q)² – 2 x q x 5 + (5)² – 25+ 21

= (q)² – 2 x q x 5 + (5)² – 25 +21

= (q)² – 2 x q x 5 + (5)² – 4

= (q – 5)² – (2)     {∵ a² – b² = (a + b) (a – b)}

= (q – 5 + 2) (q – 5 – 2)

=(q – 3) (q – 7)

问题3. 4y ² + 12y + 5

解决方案：

4y² + 12y + 5

= (2y)² + 2 x 2y x 3 + (3)² – (3)² + 5    (completing the square)

= (2y + 3)² – 9 + 5

= (2y + 3)² – 4

= (2y + 3)² – (2)²   {∵ a² – b² = (a + b) (a – b)}

= (2y + 3 + 2) (2y + 3 – 2)

= (2y + 5) (2y + 1)

问题4. p ² + 6p – 16

解决方案：

p² + 6p – 16

= (p)² + 2 x  p x 3 + (3)² – (3)² – 16    (completing the square)

= (p)² + 2 x p x 3 + (3)² – 9 – 16

= (p + 3)² – 25

= (p + 3)² – (5)²     {∵ a² – b² = {a + b) (a – b)}

= (p + 3 + 5)(p + 3 – 5)

= (p + 8) (p – 2)

问题5. x ² + 12x + 20

解决方案：

x² + 12x + 20

= (x)² + 2 x x x 6 + (6)² – (6)² + 20   (completing the square)

= (x)² + 2 x x x 6 + (6)² -36 + 20

= (x + 6)² -16

= (x + 6)² – (4)²   {∵ a² – b² = (a + b) (a – b)}

= (x + 6 + 4) (x + 6 – 4)

= (x + 10) (x + 2)

问题6. a ² – 14a – 51

解决方案：

a² – 14a – 51

= (a)² – 2 x a x 7 + (7)² – (7)² – 51       (completing the square)

= (a)² – 2 x a x 7 + (7)² – 49 – 51

= (a – 7)² – 100

= (a – 7)² – (10)²    {∵  a² – b² = (a + b) (a – b)}

= (a – 7 + 10) (a – 7 – 10)

= (a + 3) (a – 17)

问题7. a ² + 2a – 3

解决方案：

a² + 2a – 3

= (a)² + 2 x a x 1 + (1)² – (1)² – 3   (completing the square)

= (a)² + 2 x a x 1 + (1)² – 1 – 3

= (a + 1)² – 4

= (a + 1)² – (2)² {∵ a² – b² = (a + b) (a – b)}

= (a + 1 + 2) (a + 1 – 2)

= (a + 3) (a – 1)

问题8. 4x ² – 12x + 5

解决方案：

4x² – 12x + 5

= (2x)² – 2 x 2x x 3 + (3)² – (3)² + 5  (completing the square)

= (2x)² – 2 x 2x x 3 + (3)² – 9 + 5

= (2x – 3)² – 4

= (2x – 3)² – (2)²      {∵ a² – b² = (a + b) (a – b)}

= (2x – 3 + 2) (2x – 3 – 2)

= (2x – 1) (2x – 5)

问题9. y ² – 7y + 12

解决方案：

y² – 7y + 12 

= (y)² – 2 × y ×  7/2 + (7/2)² – (7/2)² + 12      (completing the square)

= (y)² – 2 × y × 7/2 + 49/4 – 49/4 + 12

= (y – 7/2)² – (49 – 48)/4 

= (y – 7/2)² – 1/4

= (y – 7/2)² – (1/2)²         {∵ a² – b² = (a + b) (a – b)}

= (y – 7/2 + 1/2) (y – 7/2 – 1/2)

= (y – 6/2)  (y – 8/2) 

= (y – 3) (y – 4)

问题10. z ² – 4z -12

解决方案：

z² – 4z – 12

= (z)² – 2 x z x 2 + (2)² – (2)² – 12  (completing the square)

= (z)² – 2 x z x 2 + (2)² – 4 – 12

= (z – 2)² – 16

= (z – 2)²– (4)²   {∵ a² – b² = (a + b) (a – b)}

= (z – 2 + 4) (z – 2 – 4)

= (z + 2)(z – 6)