Question

How do you solve \[y = 2{x^2} - 4x - 48\] by factoring?

Answer 1

Hint: We have a quadratic polynomial which is substitute to some variable ‘y’. All we need to do is find the factors of the quadratic equation and substitute the obtained factors to ‘y’. We can solve this using factorization methods or by using quadratic formulas. In factorization if it’s difficult to split the middle terms we use quadratic formula or Sridhar’s formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step answer:
Given, \[y = 2{x^2} - 4x - 48\].
Now consider the expression \[2{x^2} - 4x - 48\]. The degree of this equation is two hence we will have two factors.
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We have \[a = 2\], \[b = - 4\] and \[c = - 48\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that\[{b_1} \times {b_2} = ac\] and\[{b_1} + {b_2} = b\].
Here we can say that \[{b_1} = - 12\] and \[{b_2} = 8\]. Because \[{b_1} \times {b_2} = - 96\] \[(a \times c)\] and \[{b_1} + {b_2} = - 4(b)\].
Now we write \[2{x^2} - 4x - 48\] as,
\[ = 2{x^2} - 12x + 8x - 48\]
Taking ‘2x’ common in the first two terms and taking 8 common in the remaining two terms we have,
\[ = 2x(x - 6) + 8(x - 6)\]
Again taking \[(x - 6)\] common we have,
\[ = (x - 6)(2x + 8)\].
The factors of \[2{x^2} - 4x - 48\] are \[(x - 6)\] and \[(2x + 8)\].
Now the factors of \[y = 2{x^2} - 4x - 48\] are \[y = (x - 6)\] and \[y = (2x + 8)\].
Thus we have two linear equations and we can solve then by substitution method,
Now we have,
\[y = (x - 6)\] and \[y = (2x + 8)\]
That is
\[y - x = - 6{\text{ }} - - - (1)\]
\[y - 2x = 8{\text{ }} - - - (2)\]
From 1 we have \[y = x - 6\], substituting this in equation 2 we have,
\[x - 6 - 2x = 8\]
\[x - 2x = 8 + 6\]
\[ - x = 14\]
\[ \Rightarrow x = - 14\].
To find the ‘y’ value substitute the value of ‘x’ in equation 1 we have,
\[y - ( - 14) = - 6\]
\[y + 14 = - 6\]
\[y = - 6 - 14\]
\[ \Rightarrow y = - 20\].

Hence the solution of \[y = 2{x^2} - 4x - 48\] are \[x = - 14\] and \[y = - 20\].

Note: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts.