Question

How do you factor the expression ${y^2} + 6y - 16 = 0$?

Answer 1

Hint: First we will reduce the equation further if possible. Then we will try to factorise the terms in the equation. Then solve the equation by using the quadratic formula and finally evaluate the value of the variable accordingly.

Complete step-by-step solution:
We will start off by reducing any reducible terms in the equation if possible.
${y^2} + 6y - 16 = 0$
Now we will factorise the terms in the equation.
${y^2} + 6y - 16 = 0$
Now we will try to factorise by using the quadratic formula which is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
To substitute the values, we first compare and evaluate the values from the general form of quadratic equation. the general form of quadratic equation is given by $a{x^2} + bx + c = 0$.
When we compare the terms, we get the values as,
$
  a = 1 \\
  b = 6 \\
  c = - 16 \\
 $
Now substitute all these values in the quadratic formula, to evaluate the value of the variable.
\[
\Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow y = \dfrac{{ - (6) \pm \sqrt {{{(6)}^2} - 4(1)( - 16)} }}{{2(1)}} \\
\Rightarrow y = \dfrac{{ - (6) \pm \sqrt {(36) - 4(1)( - 16)} }}{{2(1)}} \\
\Rightarrow y = \dfrac{{ - 6 \pm \sqrt {(36) + 64} }}{2} \\
\Rightarrow y = \dfrac{{ - 6 \pm \sqrt {100} }}{2} \\
\Rightarrow y = \dfrac{{ - 6 \pm 10}}{2} \\
\Rightarrow y = - 3 \pm 5 \\
 \]
Now we solve for the value $y$ separately.
So, we get the values as,
\[
\Rightarrow x = - 3 - 5 = - 8 \\
\Rightarrow x = - 3 + 5 = 2 \\
 \]
Hence, the total valid solutions of the quadratic equation ${y^2} + 6y - 16 = 0$ are \[ - 8,2\].
Additional information: The quadratic formula is the formula that provides the solution to a quadratic equation. there are two ways of solving a quadratic equation instead of using the quadratic formula, such as factoring, completing the square, graphing and others.

Note: While splitting the middle term be careful. After splitting the middle term Do not solve all the equations simultaneously. Solve all the equations separately, so that you don’t miss any term of the solution. Check if the solution satisfies the original equation completely. If any term of the solution doesn’t satisfy the equation, then that term will not be considered as a part of the solution.