Question

Multiply. start fraction k plus 3 over 4 k minus 2 end fraction dot left parenthesis 12 k squared plus 2 k minus 4 right parenthesis

Answer 1

`\therefore ((k+3))/((4k-2)).(12k^2+2k-4)=3k^2+11k+6`

Explanation:

Factorization of a Quadratic polynomial:

• In order to factorize
  `ax^2+bx+c` we have to find out the numbers p and q such that, p+q = b and pq=ac.

• Finding the two integers p and q, we rewrite the middle term of the quadratic as px+qx. Then by grouping of the terms we can get desired factors.

Multiplication of two binomial:

(a+b)(c+d)

=a(c+d)+b(c+d)

=(ac+ad)+(bc+bd)

=ac+ad+bc+bd

Given that,

`((k+3))/((4k-2)).(12k^2+2k-4)`

`=((k+3))/(2(2k-1)).2(6k^2+k-2)` [ taking common 2]

`=((k+3))/((2k-1)).(6k^2+k-2)` [ cancel 2]

`=((k+3))/((2k-1)).(6k^2+4k-3k-2)`

`=((k+3))/((2k-1)).\{2k(3k+2)-1(3k+2)\}`

`=((k+3))/((2k-1)).(3k+2)(2k-1)`

`=(k+3).(3k+2)`

`=3k^2+9k+2k+6`

`=3k^2+11k+6`

`\therefore ((k+3))/((4k-2)).(12k^2+2k-4)=3k^2+11k+6`