Help plss and explain!! If 7x^2 - mx-12 is equal to (7x + n)(x-6), where m and n are constants, find the value of m.

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

jml angle lkj & angle mj = lm

the answer is -1 1/5, also a

hope i

-lavira : )