2x(3x + 5) + 3(3x + 5) = ax² + bf + c
2x.3x + 2x.5 + 3.3x + 3.5 = ax² + bf + c
6x² + 10x + 9x + 15 = ax² + bf + c
6x² + 19x + 15 = ax² + bf + c
6x² = ax²
6 = a
19x = bf
15 = c
As the question says, b is constant, so he can't be a variable, so he only can be 19 and f be x
19x = bf
19 = b
x = f
The value of b is 17. The value can be derived as shown below.
[tex]2x(3x+5)+3(3x+5)=ax^2+bx+c[/tex]
The equation is true for all values of x.
Since it is given that the equation is true for all values of x, thus:
At x = 0:
[tex]2x(3x+5)+3(3x+5)=ax^2+bx+c\\\\2(0)(3(0) + 5) + 3(3(0) + 5) = a(0)^2 + b(0) + c\\\\15 = c[/tex]
Thus, we get c = 15
Now putting x = 1:
[tex]2x(3x+5)+3(3x+5)=ax^2+bx+c\\\\2(1)(3(1) + 5) + 3(3(1) + 5) = a(1)^2 + b(1) + 15\\\\40 = a + b + 15\\a + b = 25[/tex]
Putting x = -1:
[tex]2x(3x+5)+3(3x+5)=ax^2+bx+c\\\\2(-1)(3(-1) + 5) + 3(3(-1) + 5) = a(-1)^2 + b(-1) + 15\\\\2 = a - b + 15\\a - b = -13[/tex]
Thus we have 2 equations:
[tex]a + b = 25\\a - b = -13[/tex]
Subtracting second equation from first equation we get:
[tex]b + b = 25 + 13\\\\b = \dfrac{38}{2}\\\\b = 17[/tex]
Thus, the value of b is 17.
Learn more about linear equations here:
https://brainly.com/question/14362668
