Question

Solve the following quadratic equations by factorization:ax²+(4a²-3b)x-12ab=0

Answer 1

The solved given quadratic equation by using the Factorization method is (ax-3b)(x+4a)

Step-by-step explanation:

Given quadratic equation is $ax^2+(4a^2-3b)x-12ab=0$

To solve the given quadratic equation by Factorization method :

• $ax^2+(4a^2-3b)x-12ab=0$
• $ax^2+4a(x)-3b(x)-12ab=0$ ( here using the distributive property (x-y)z=xz-yz where x=$4a^2$ , y=3b and z=x )
• $ax^2+4ax-3bx-12ab=0$
• $ax(x+4a)-3b(x+4a)=0$ ( taking the common term outside the factors )
• $(ax-3b)(x+4a)=0$ ( taking the common factor outside the factors )
• Therefore $ax^2+(4a^2-3b)x-12ab=(ax-3b)(x+4a)=0$

Therefore the solved given quadratic equation by using the Factorization method is (ax-3b)(x+4a)