+0  
 
-2
31
2
avatar+817 

The expression $6x^2 + 17x + 5 - 5x^2 - 11x$ can be written in the form $(Ax+1)(Bx+5)$ where $A$ and $B$ are integers. What is the value of $AB$?

 Aug 8, 2023
 #1
avatar
0

To factor the given expression $6x^2 + 17x + 5 - 5x^2 - 11x$, let's combine the like terms first:

$6x^2 + 17x + 5 - 5x^2 - 11x = (6x^2 - 5x^2) + (17x - 11x) + 5 = x^2 + 6x + 5$

Now we need to factor the expression $x^2 + 6x + 5$ into the form $(Ax+1)(Bx+5)$. We'll look for integers $A$ and $B$ that satisfy this factorization.

To do this, we need to find two numbers that multiply to the coefficient of the constant term (which is 5) and add up to the coefficient of the linear term (which is 6). These numbers are 5 and 1, because:

$5 \times 1 = 5$ (constant term) $5 + 1 = 6$ (linear term)

So, we can factor $x^2 + 6x + 5$ as:

$x^2 + 6x + 5 = (x + 5)(x + 1)$

Thus, $A = 1$ and $B = 5$.

The value of $AB$ is $A \times B = 1 \times 5 = 5$.

 Aug 8, 2023
 #2
avatar+129771 
+1

Simplify the polynomial as

 

x^2  + 6x + 5

 

This can be factored as   (x+5) (x +1)

 

A,B  = 1

 

AB = (1)(1)  =  1

 

 

cool cool cool

 Aug 8, 2023
edited by CPhill  Aug 8, 2023

0 Online Users