Home
Class 9
MATHS
7(x-2y)^2 - 25 (x-2y)+12...

`7(x-2y)^2 - 25 (x-2y)+12`

Text Solution

Verified by Experts

Solution
Given: `7(x-2y)2 - 25(x-2y) + 12`
Let us consider `x - 2y = z`
So we get,
= `7z²-25z + 12`
We can further write it as
= `7z²-21z4z + 12`
By taking out the common terms = `7z (z-3) -4 (z-3)` = `(7z-4) (z-3)`
Promotional Banner

Topper's Solved these Questions

  • EXPONENTS OF REAL NUMBER

    RD SHARMA|Exercise All Questions|186 Videos

  • FACTORIZATION OF POLYNOMIAL

    RD SHARMA|Exercise All Questions|220 Videos

Similar Questions

Explore conceptually related problems

Factorise: 25(x+y)^(2)- 36 (x-2y)^(2).

Factor (x + y) ^ (2) -7 (x ^ (2) -y ^ (2)) + 12 (xy) ^ (2)

If x^2+y^2 = 25 and xy = 12, then the value of x^(-1)+y^(-1) is :

Simplify : ( 3y ( x - y) - 2x ( y -2x))/( 7x ( x - y) - 3 ( x ^(2) - y ^(2)))

Simplify: 25(2x+y)^(2)-16(x-y)^(2)

{:("Column" A ,, "Column" B), (225x^(2) - 625 y^(2) = ,, (a) 25(x-2) (x-2)), (x^(2) - x - y - y^(2) = ,, (b) 25(3x- 5y) (3x + 5y)), (x^(2) - x - y^(2) + y = ,, (x + y) (x - y- 1)), (25x^(2) - 100 x + 100 = ,, (d) (x - y) (x + y -1)), (,,(e) (x + y) (x + y - 1)):}

Factorize each of the following algebraic expressions: 6x(2x-y)7y(2x-y)2r(y-x)+s(x-y)7a(2x-3)+3b(2x-3)9a(6a-5b)-12a^(2)(6a-5b)

Find 'all equation to the four common tangents to the circles x^(2)+y^(2)=25 and (x-12)^(2)+y^(2)=9