Question

How do you solve \[4{x^2} - 17x - 15 = 0\] 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. If factors are difficult to find then we use Sridhar’s formula to find the roots. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step solution:
Given, \[4{x^2} - 17x - 15 = 0\]
Since the degree of the equation is 2, we have 2 factors.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 4\], \[b = - 17\] and \[c = - 15\].
The standard form of the factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c = 0\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
Thus we will have \[4{x^2} - 20x + 3x - 15 = 0\] which satisfies the condition \[\left( { - 20} \right) \times \left( 3 \right) = - 60\left( {a \times c} \right)\] and \[\left( { - 20} \right) + 3 = - 17\left( b \right)\].
Then we have
\[ \Rightarrow 4{x^2} - 17x - 15 = 4{x^2} - 20x + 3x - 15\]
\[ = 4{x^2} - 20x + 3x - 15\]
Now taking ‘4x’ common in the first two terms and taking 3 common in the remaining two terms we will have
\[ = 4x(x - 5) + 3(x - 5)\]
Now taking \[(x - 5)\] common we will have,
\[ = \left( {4x + 3} \right)(x - 5)\]
Thus we have two factors.
Now to find the roots,
\[\left( {4x + 3} \right)(x - 5) = 0\]
By using the zero product principle we have,
\[\left( {4x + 3} \right) = 0\] or \[(x - 5) = 0\].
\[4x = - 3\] or \[x = 5\].
\[ \Rightarrow x = - \dfrac{3}{4}\] or \[x = 5\].
This is the required answer.
Additional information:
We can also solve this using the quadratic formula. That is,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting the known values we have,
\[ \Rightarrow x = \dfrac{{ - ( - 17) \pm \sqrt {{{( - 17)}^2} - 4(4)( - 15)} }}{{2(4)}}\]
\[x = \dfrac{{17 \pm \sqrt {289 + 240} }}{8}\]
(Because the product of two negative number results in positive number)
\[x = \dfrac{{17 \pm \sqrt {529} }}{8}\]
We know 529 is a perfect square,
\[x = \dfrac{{17 \pm 23}}{8}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{17 + 23}}{8}\] and \[x = \dfrac{{17 - 23}}{8}\]
\[ \Rightarrow x = \dfrac{{40}}{8}\] and \[x = \dfrac{{ - 6}}{8}\]
\[ \Rightarrow x = 5\] and \[x = \dfrac{{ - 3}}{4}\]. This is the required answer.

Note: These are the roots of the given polynomial of degree 2. In above, 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. Careful in the calculation part.