Answer:
[tex](x-3)(-5x-2)[/tex]
or [tex]-(x-3)(5x+2)[/tex]
Step-by-step explanation:
quadratic equation format: [tex]a^2+b^2+c[/tex]
Therefore, for [tex]-5x^2+13x+6[/tex], a = -5, b = 13 and c = 6
Multiply the coefficient of [tex]x^2[/tex] ([tex]a[/tex]) by the constant term ([tex]c[/tex])
[tex]\implies -5 \times 6 = -30[/tex]
Find two numbers which have a product of -30 and a sum of [tex]b[/tex] (13)
Factors of 30: 1 and 30, 2 and 15, 3 and 10, 5 and 6
So -2 and 15 have a product of -30 and sum of 13.
Rewrite [tex]13x[/tex] as [tex]15x-2x[/tex] and substitute into the equation:
[tex]\implies -5x^2+15x-2x+6[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies-5x(x-3)-2(x-3)[/tex]
The bracket created should always be the same.
The two brackets have now been found. The first bracket is the common factor of (x - 3). The second bracket is the factorized terms outside of each bracket (-5x - 2).
[tex]\implies (x-3)(-5x-2)[/tex]
Additionally, we can write (-5x - 2) as -(5x + 2)
[tex]\implies -(x-3)(5x+2)[/tex]
