Question

If the quadratic 3x^2+bx+10 can be written in the form a(x+m)^2+n, where m and n are integers, what is the largest integer that must be a divisor of b?

Answer 1

Answer:

6

Step-by-step explanation:

3x² + bx + 10 = 3(x² + $\frac{b}{3}$ x) + 10 =

3(x² + 2 × $\frac{b}{6}$ x + ($\frac{b}{6}$)² - ($\frac{b}{6}$)²) + 10 =

3[(x + $\frac{b}{6}$ )² - ( $\frac{b}{6}$ )²] + 10 =

3(x + $\frac{b}{6}$ )² - 3( $\frac{b}{6}$ )² + 10

Now it is in the form a(x + m)² + n where

a = 3, m = $\frac{b}{6}$ , n = - 3( $\frac{b}{6}$ )² + 10

In order for "m" to be an integer, "b" must be a multiple of 6.

In order for "n" to be an integer, "b" must be a multiple of 6.

The largest integer that must be a divisor of "b" in order for both "m" and "n"  to be integers is 6.  

Answer 2

Given : quadratic 3x^2+bx+10 can be written in the form a(x+m)^2+n, where m and n are integers

To find :  largest integer that must be a divisor of b

Solution:

3x²+bx+10

a(x+m)²+n

= a(x² + m² + 2xm) + n

= ax²  + 2amx +  am² + n

= ax² + 2amx + ( am² + n)

Equating  with  3x²+bx+10  

a = 3  

2am  = b

=> 2(3)m = b

=> m = b/6

=> b = 6m

Hence b must have a divisor of  6  

6 is the integer  that must be a divisor of b

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