Question

How do you factor \[{{p}^{2}}+18t+32\]?

Answer 1

Hint: If \[x=a\] is a root of a polynomial function, then \[x-a\] is one of its factors. To express a quadratic equation \[a{{x}^{2}}+bx+c\] in its factored form. We have to find its roots, say \[\alpha ,\beta \] are the two real roots of the equation. Then the factored form is \[a\left( x-\alpha \right)\left( x-\beta \right)\]. We can find the roots of the equation using the formula method as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step solution:
We are given the quadratic expression \[{{p}^{2}}+18t+32\]. On comparing with the general solution of the quadratic equation \[a{{x}^{2}}+bx+c\], we get \[a=1\].
To express in factored form, we first have to find the roots of the equation \[{{t}^{2}}+8t+12\].
We can find the roots of the equation using the formula method.
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of the coefficients in the above formula, we get
\[\begin{align}
  & \Rightarrow p=\dfrac{-(18)\pm \sqrt{{{\left( 18 \right)}^{2}}-4(1)(32)}}{2(1)} \\
 & \Rightarrow p=\dfrac{-18\pm \sqrt{196}}{2} \\
 & \Rightarrow p=\dfrac{-18\pm 14}{2} \\
\end{align}\]
\[\Rightarrow p=\alpha =\dfrac{-18+14}{2}=-2\] or \[p=\beta =\dfrac{-18-14}{2}=-16\]
Now that, we have the roots of the given expression, we can express it as its factored form as follows,
For the quadratic expression \[{{p}^{2}}+18t+32\], \[a=1\] and the roots as \[\alpha =-2\And \beta =-16\].
The factored form is,
\[\begin{align}
  & \Rightarrow 1\left( p-\left( -2 \right) \right)\left( x-\left( -16 \right) \right) \\
 & \Rightarrow \left( p+2 \right)\left( p+16 \right) \\
\end{align}\]

Note: As we use the roots of the expression, to make the factored form. We can also use the factored form to find the roots of an expression.
For example, if we are given the expression \[\left( t+2 \right)\left( t+6 \right)\]. As the degree of the expression, we can get by expanding these brackets is 2. We can say that the quadratic expression formed by these factors has two real roots and they are \[t+2=0\And t+6=0\]. On solving these equations, we get \[t=-2\And t=-6\]. These are the roots of the given quadratic equation.