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

Answer:

$\therefore \frac{(k+3)}{(4k-2)}.(12k^2+2k-4)=3k^2+11k+6$

Step-by-step 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,

$\frac{(k+3)}{(4k-2)}.(12k^2+2k-4)$

$=\frac{(k+3)}{2(2k-1)}.2(6k^2+k-2)$       [ taking common 2]

$=\frac{(k+3)}{(2k-1)}.(6k^2+k-2)$          [ cancel 2]

$=\frac{(k+3)}{(2k-1)}.(6k^2+4k-3k-2)$

$=\frac{(k+3)}{(2k-1)}.\{2k(3k+2)-1(3k+2)\}$

$=\frac{(k+3)}{(2k-1)}.(3k+2)(2k-1)$

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

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

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

$\therefore \frac{(k+3)}{(4k-2)}.(12k^2+2k-4)=3k^2+11k+6$