X^2+13x+42=(x+6)(x-a)

Предыдущий
вопрос Следующий
вопрос

To find the value of 'a', expand the right side of the equation:

(x+6)(x-a) = x^2 - ax + 6x - 6a
= x^2 - (a-6)x - 6a

Now, we can set the expanded form equal to the original equation:

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

Now we can compare the coefficients on both sides:

13 = -(a-6) (coefficients of x)
42 = -6a (constant term)

From first equation, we get:
13 = -a + 6
a = 6 - 13
a = -7

So, the value of 'a' is -7.