(i) (x + 2y + 3) (x + 2y + 7)
(ii) (2x + y + 5) (2x + y – 9)
(iii) (x – 2y – 5) (x – 2y + 3)
(iv) (3x – 4y – 2) (3x – 4y – 6)
Answer :
(i) (x + 2y + 3) (x + 2y + 7)
Consider x + 2y = a
(a + 3) (a + 7) = a2 + (3 + 7) a + 3 × 7
By further calculation
= a2 + 10a + 21
Substituting the value of a
= (x + 2y)2 + 10 (x + 2y) + 21
By expanding using formula
= x2 + 4y2 + 2 × x × 2y + 10x + 20y + 21
So we get
= x2 + 4y2 + 4xy + 10x + 20y + 21
(ii) (2x + y + 5) (2x + y – 9)
Consider 2x + y = a
(a + 5) (a – 9) = a2 + (5 – 9) a + 5 × (-9)
By further calculation
= a2 – 4a – 45
Substituting the value of a
= (2x + y)2 – 4 (2x + y) – 45
By expanding using formula
= 4x2 + y2 + 2 × 2x × y – 8x – 4y – 45
So we get
= 4x2 + y2 + 4xy – 8x – 4y – 45
(iii) (x – 2y – 5) (x – 2y + 3)
Consider x – 2y = a
(a – 5) (a + 3) = a2 + (- 5 + 3) a + (-5) (3)
By further calculation
= a2 – 2a – 15
Substituting the value of a
= (x – 2y)2 – 2 (x – 2y) – 15
By expanding using formula
= x2 + 4y2 – 2 × x × 2y – 2x + 4y – 15
So we get
= x2 + 4y2 – 4xy – 2x + 4y – 15
(iv) (3x – 4y – 2) (3x – 4y – 6)
Consider 3x – 4y = a
(a – 2) (a – 6) = a2 (- 2 – 6)a + (-2) (-6)
By further calculation
= a2 – 8a + 12
Substituting the value of a
= (3x – 4y)2 – 8 (3x – 4y) + 12
Expanding using formula
= 9x2 + 16y2 – 2 × 3x × 4y – 24x + 32y + 12
So we get
= 9x2 + 16y2 – 24xy – 24x + 32y + 12
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