Respuesta :
Answer:
[tex]\therefore \frac{(k+3)}{(4k-2)}.(12k^2+2k-4)=3k^2+11k+6[/tex]
Step-by-step explanation:
Factorization of a Quadratic polynomial:
- In order to factorize [tex]ax^2+bx+c[/tex] 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,
[tex]\frac{(k+3)}{(4k-2)}.(12k^2+2k-4)[/tex]
[tex]=\frac{(k+3)}{2(2k-1)}.2(6k^2+k-2)[/tex] [ taking common 2]
[tex]=\frac{(k+3)}{(2k-1)}.(6k^2+k-2)[/tex] [ cancel 2]
[tex]=\frac{(k+3)}{(2k-1)}.(6k^2+4k-3k-2)[/tex]
[tex]=\frac{(k+3)}{(2k-1)}.\{2k(3k+2)-1(3k+2)\}[/tex]
[tex]=\frac{(k+3)}{(2k-1)}.(3k+2)(2k-1)[/tex]
[tex]=(k+3).(3k+2)[/tex]
[tex]=3k^2+9k+2k+6[/tex]
[tex]=3k^2+11k+6[/tex]
[tex]\therefore \frac{(k+3)}{(4k-2)}.(12k^2+2k-4)=3k^2+11k+6[/tex]
