Question

Factorisation
6. (i) 6x2 + 11x - 10
(ii) 5 – 4x – 12
- 35
y – 12y2
21pq +9
7) - 10
-- 10.]
(ii) 8 + 6(p + q) - 5(p +
- 5(+q)²​

Answer 1

Answer:1.Ur solution here!

➡ 6x^2 +11x -10 =0

➡ 6x^2 +15x -4x -10 =0

➡ 3x (2x +5) -2(2x +5) =0

➡ (3x -2) (2x+5) =0

X = 2/3 & -5/2

2.Step-1 : Multiply the coefficient of the first term by the constant   12 • -35 = -420  

Step-2 : Find two factors of  -420  whose sum equals the coefficient of the middle term, which is   -1 .

    -420   +   1   =   -419

    -210   +   2   =   -208  

  -140   +   3   =   -137

    -105   +   4   =   -101    

 -84   +   5   =   -79  

   -70   +   6   =   -64  

  -60   +   7   =   -53    

 -42   +   10   =   -32

    -35   +   12   =   -23

    -30   +   14   =   -16  

   -28   +   15   =   -13  

   -21   +   20   =   -1  

 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -21  and  20  

                    12x2 - 21x + 20x - 35

Step-4 : Add up the first 2 terms, pulling out like factors :

                   3x • (4x-7)

             Add up the last 2 terms, pulling out common factors :

                   5 • (4x-7)

Step-5 : Add up the four terms of step 4 :

                   (3x+5)  •  (4x-7)

            Which is the desired factorization

Final result : (4x - 7) • (3x + 5)

.

3.Let p+q=a---(1)

8+6(p+q)-5(p+q)^2

=8+6a-5a^2

=8-4a+10a-5a^2

=4(2-a)+5a(2-a)

=(2-a)(4+5a)

From(1)

=[2-(p+q)][4+5(p+q)]

hope it helps u

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=(2-p-q)(4+5p+5q)

Step-by-step explanation: