If x^2 - 13x + 42 = (x + A)(x + B), then A + B =?

Question

If x^2 - 13x + 42 = (x + A)(x + B), then A + B =

-42
-1
13
42
none of these

Answer 1

(x-6)(x-7)

the answer is none of these

it would be -13

Answer 2

x^2 + 13x + 40 2 Set this equivalent to 0, and then we are going to locate both roots for this polynomial. x^2 + 13x + 40 2 = 0 all of us keep in mind that x cases x supplies us x^2, yet we go with to carry close aspects of 40 2. there is two * 21, 6 * 7, 3 * 14... we go with to apply both aspects, that once extra jointly, supply us 13. 6 + 7 = 13, so those are the aspects we are going to use. (x + 6)(x + 7) = 0. Now we set both one in each and every of those equivalent to 0. x + 6 = 0 and x + 7 = 0. We subtract 6 from each and every part, and 7 from each and every part, and we get: x = -6, and x = -7

Answer 3

13

Answer 4

x^2 - 13x + 42
=(x - 7)( x -6)

-7 + (-6) = -13

So, none of these

Answer 5

the right hand side becomes
x^2 + (A+B)x + AB.
that means, by equating coefficients (or what's infront of each DIFFERENT x^whatever term)
A+B = -13 (eq 1)
AB = 42. (eq 2)
use eq 1 to rewrite B in terms of A and plug that into the 2nd equation to solve for B. then use eq 1 and what B is to find A.

Answer 6

x^2-13x+42 = (x-6)(x-7) = (x+A)(x+B)

A = -6, and B = -7.

-6 + -7 = -13

The answer is None of These.

Answer 7

A and B are the roots of the equation.
we know that the sum of the roots is -b/a and the product of the roots is c/a.

A + B = -b/a = 13/1 = 13

Answer 8

x²-13x+42=(x+A)(x+B)
x²-13x+42=x²+(A+B)x+AB
A+B=-13; AB=42
none of these

Answer 9

-13

Answer 10

x² - 13x + 42 = (x - 7)(x - 6) = (x + A)(x + B)
A = - 7
B = - 6
A + B = - 13
NONE of THESE