Question

Factor completely the trinomial, if it is factorable using integers enter nf7 - 48x - 7x^2

Answer 1

`\text{Given: }7-48x-7x^2`

Since the given trinomial is a quadratic expression. We can factor this using a quadratic formula.

`x=( -b \pm√(b^2 - 4ac))/( 2a )`
`\begin{gathered} \text{Rearrange the expression so that it is in the form }ax^2+bx+c \\ \\ 7-48x-7x^2\Rightarrow-7x^2-48x+7 \\ \\ \text{Here, we can determine }a,b,\text{ and }c \\ a=-7 \\ b=-48 \\ c=7 \end{gathered}`

Substitute the following coefficients the quadratic formula and we get:

`\begin{gathered} x=( -b \pm√(b^2 - 4ac))/( 2a ) \\ x=( -(-48) \pm√((-48)^2 - 4(-7)(7)))/( 2(-7) ) \\ x=\frac{48\pm\sqrt[]{2304-(-196)}}{-14} \\ x=( 48 \pm√(2500))/( -14 ) \\ x=( 48 \pm50\, )/( -14 ) \\ x=-( 98 )/( 14 )\; \; \; x=( 2 )/( 14 ) \\ x=-7\; \; \; x=( 1)/( 7 ) \end{gathered}`

Equate the solution to zero to get the factors.

`\begin{gathered} x=-7 \\ x+7=0 \\ \\ x=(1)/(7) \\ x-(1)/(7)=0 \\ 7x-1=0 \\ \\ \text{Therefore, the factor of }7-48x-7x^2\text{ is }(x+7)(7x-1) \end{gathered}`