Question

using the standard form, (MATH PROBLEM) of the group given quadratic equation, factor the left hand side of the equation into two linear factors

Answer 1

Factors are (6x-5) and (7x+1).

Given:

The quadratic equation is

$42x^2=29x+5$

The objective is to factor the equation.

The quadratic equation can be written as,

$\begin{gathered} 42x^2=29x+5 \\ 42x^2-29x-5=0 \end{gathered}$

On factorizing the above equation,

$\begin{gathered} 42x^2+6x-35x-5=0 \\ 6x(7x+1)-5(7x+1)=0 \\ (6x-5)(7x+1)=0 \end{gathered}$

Hence, the factors are (6x-5)(7x+1)=0.