Question

How do you solve \[8{m^2} + 20m = 12\] by factoring?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. Here we have ‘m’ instead of ‘x’ as variable. If factors are difficult to find then we use Sridhar’s formula to find the roots. That is \[m = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[8{m^2} + 20m = 12\]
Rearranging the equation we have,
\[8{m^2} + 20m - 12 = 0\]
Since the degree of the equation is 2, we have 2 roots.
On comparing the given equation with the standard quadratic equation\[a{m^2} + bm + c = 0\], we have\[a = 8\], \[b = 20\] and \[c = - 12\].
The standard form of the factorization of quadratic equation is \[a{m^2} + {b_1}m + {b_2}m + c = 0\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
We can write the given equation as \[8{m^2} + 24m - 4m - 12 = 0\], where \[{b_1} = 24\] and \[{b_2} = - 4\]. Also \[{b_1} \times {b_2} = 24 \times ( - 4) = - 96(ac)\] and \[{b_1} + {b_2} = 24 - 4 = 20(b)\].
Thus we have,
\[ \Rightarrow 8{m^2} + 20m - 12 = 8{m^2} + 24m - 4m - 12\]
\[ = 8{m^2} + 24m - 4m - 12\]
Taking ‘8x’ common in the first two terms and taking -4 common in the remaining two terms we have,
\[ = 8m(m + 3) - 4(m + 3)\]
Again taking \[(m + 3)\] common we have,
\[ = (8m - 4)(m + 3)\]
Hence the factors of \[8{m^2} + 20m = 12\] are \[(8m - 4)\] and \[(m + 3)\].
Now to find the roots
\[(8m - 4)(m + 3) = 0\]
By zero product principle we have,
\[ \Rightarrow (8m - 4) = 0\] and \[(m + 3) = 0\]
\[ \Rightarrow 8m = 4\] and \[m = - 3\]
\[ \Rightarrow m = \dfrac{4}{8}\] and \[m = - 3\].
\[ \Rightarrow m = \dfrac{1}{2}\] and \[m = - 3\]. These are the solutions of \[8{m^2} + 20m = 12\].

Note: If we are unable to expand the middle term of the given equation into a sum of two numbers then we use a quadratic formula to solve the given problem. Quadratic formula and Sridhar’s formula are both the same. If a polynomial is of degree ‘n’ then we have ‘n’ roots. Here the degree of the polynomial is 2 hence we have 2 roots. Careful in the calculation part.