Answer:
m=40
Problem:
If 7x^2 - mx-12 is equal to (7x + n)(x-6), where m and n are constants, find the value of m.
Step-by-step explanation:
We want to find n amd m such that
(7x+n)(x-6)=7x^2-mx-12.
Since (ax+b)(cx+d)=acx^2+(ad+bc)x+bd, then we need or should have the following:
(7x)(x)=7x^2
(7×-6+n×1)x=-mx
(n)(-6)=-12
The bottom equation tells us n=2 since 2(-6)=-12.
The first equation is already true.
Now we must solve (7×-6+n×1)x=-mx with n=2 for m.
That is we need to solve 7×-6+2×1=-m
Simplify and done -m=-42+2=-40 so m=40.
Let's do a check
7x^2 - mx-12 is equal to (7x + n)(x-6)
7x^2-40x-12 is equal to (7x+2)(x-6)
(7x+2)(x-6)=7x(x)+7x(-6)+2(x)+2(-6)
(7x+2)(x-6)=7x^2-42x+2x-12
(7x+2)(x-6)=7x^2-40x-12 and that is what we wanted.
Another way:
We want (7x+n)(x-6)=7x^2-mx-12 to be true for all x.
So if x=0 or for x=1, we want the equation to be true.
Insert x=0. This gives (n)(-6)=-12 which implies n=2.
(7x+2)(x-6)=7x^2-mx-12
Insert x=1. This gives (7+2)(1-6)=7-m-12.
Simplify both sides: 9(-5)=-m-5
Continue to simplify left side: -45=-m-5
Add 5 on both sides: -40=-m
Multiply both sides by -1: 40=m
