Question

6 - 5 factoring ( ax^2 + bx + c )
5 2 = 10? 5 and 2
5 + 2 = 7
quadratic expression ( ax^2 + bx + c )
( a cdot c = ? )
( square \times square )
( square + square = b )

1. ( 7x^2 - 18x - 9 )

( a cdot c = 7 \times (-9) = -63 )
( square cdot square = -63 )
( square + square = -18 )
( (x - 3)(7x + 3) ) (boxed in pink)
( 7x^2 ), ( -21x )
( gcf = 7 )
( x^2 ), ( gcf = 7 )
( -9 ), ( -21x )
( gcf = 3 )
( 3x ), ( gcf = 1 )
( 7x^2 ), ( gcf = 7 )
(also includes a box - method diagram with ( 7x^2 ), ( -21x ), ( 3x ), ( -9 ) and factoring steps with gcf (greatest common factor) markings)

Explanation:

Step1: Rewrite the quadratic equation

First, rewrite the quadratic expression \(7x^{2}-18x - 9\) in standard form \(ax^{2}+bx + c\) (it is already in standard form with \(a = 7\), \(b=- 18\), \(c=-9\)). We need to find two numbers that multiply to \(a\times c=7\times(-9)=- 63\) and add up to \(b = - 18\).
The two numbers are \(-21\) and \(3\) since \(-21\times3=-63\) and \(-21 + 3=-18\).

Step2: Split the middle term

Split the middle term \(-18x\) using the two numbers we found:
\(7x^{2}-21x + 3x-9\)

Step3: Factor by grouping

Group the first two terms and the last two terms:
\((7x^{2}-21x)+(3x - 9)\)
Factor out the greatest common factor (GCF) from each group. From the first group, the GCF of \(7x^{2}\) and \(-21x\) is \(7x\), so we get \(7x(x - 3)\). From the second group, the GCF of \(3x\) and \(-9\) is \(3\), so we get \(3(x - 3)\).
Now we have \(7x(x - 3)+3(x - 3)\).

Step4: Factor out the common binomial factor

Notice that both terms have a common factor of \((x - 3)\). Factor that out:
\((x - 3)(7x + 3)\)

Answer:

The factored form of \(7x^{2}-18x - 9\) is \(\boldsymbol{(x - 3)(7x + 3)}\)