Question

3. Explain how a four-term polynomial is factored by grouping and when a quadratic trinomial can be factored using this method. Include examples in your explanation

Answer 1

look, inkyvoyd has given you all of the "how too" and step by steps for solving up to here, but the below equation is a similar example, for reference, if it also helps?
\[\large x^2+bx+20\]

use FOIL to factorise
\[\large (x+J)(x+K)\]

so\[\large J × K=20\]
and \[\large J + K=b\]


so possible answers for J and K are 4 and 5, 2 and 10, 1 and 20

lets use 4 and 5, jus coz
proof:
\[\large (x+5)(x+4)\]
now expand using FOIL
First = \[\large x×x=x^2\]

Outer = \[\large x×4=4x\]

Inner = \[\large 5×x=5x\]

Last = \[\large 5×4=20\]

so
full equation if we use this is
\[\large x^2+4x+5x+20 →x^2+9x+20\]

so b = 9x, and is made up of the terms 4x and 5x

follow so far?

now, if that equation is changed to :
\[\large 2x^2+bx+20\]
... using FOIL, we still factorise to:
\[\large (x+L)(2x+M)\]

so\[\large L × M=20\]

and \[\large 2L + M=b →(\text {as right bracket is 2x + M})\]


so possible answers for L and M are 4 and 5, 2 and 10, 1 and 20
lets use 4 and 5 again, just coz
proof:
\[\large (x+5)(2x+4)\]
now expand using FOIL
First = \[\large x×2x=2x^2\]

Outer = \[\large x×4=4x\]

Inner = \[\large 5×2x=10x\]

Last = \[\large 5×4=20\]

so
full equation if we use this is
\[\large 2x^2+4x+10x+20 →2x^2+14x+20\]

so b = 14x, and is made up of the terms 4x and 10x

does this example help at all? @BasedGod1122 ???

Answer 2

alternatively, if b in \[x^2+bx+20\]
= b in
\[2x^2+bx+20\]

then we sub in our original b value of 9, and solve 2nd equation \[2x^2+bx+20\] using quadratic formula