Question

If (x - a) is a factor of (3x² - mx - nx), then prove that (a = m + n/3).

Answer 1

`\text{Since }(x-a)\text{ is a factor of }3x^2-mx-nx,\text{ putting }x=a\text{ in the expression}\\\text{makes the expression equal to zero.}\\\therefore\ 3a^2-m(a)-n(a)=0\\\text{or, }3a^2-a(m+n)=0\\\text{or,\ }a\{3a-(m+n)\}=0\\\text{i.e. }a=0\text{ or }3a-(m+n)=0\\\text{Taking the second case,}\\3a-(m+n)=0\\\text{or,\ }3a=m+n\\\therefore\ a=(m+n)/(3)`

Answer 2

By applying the Factor Theorem, we proved that if (x - a) is a factor of (3x² - mx - nx), then a must equal (m + n/3), since substituting x with a makes the polynomial equal zero and we can isolate a to find its value.

Step-by-step explanation:

To prove that (a = m + n/3) given that (x - a) is a factor of (3x² - mx - nx), we can start by applying the Factor Theorem. According to the theorem, if (x - a) is a factor, then the polynomial should equal zero when we substitute x with a. Therefore, the equation becomes:

3a² - ma - na = 0.

To find a, we rearrange to solve for a:

a(3a - m - n) = 0.

Since we're interested in the case when a is not zero, we have:

3a - m - n = 0.

Now, we isolate a:

a = (m + n) / 3.

So, we have proven that if (x - a) is a factor of (3x² - mx - nx), then (a = m + n/3).