Question

Разложите на множители квадратный трёхчлен: x2+x -12; x2+3x -10; 3a2+7a -6;

Answer 1

1) x² + x - 12 = 0

По теореме Виета :

x₁ = - 4      x₂ = 3

x² + x - 12 = (x + 4)(x - 3)

2) x² + 3x - 10 = 0

По теореме Виета :

x₁ = - 5      x₂ = 2

x² + 3x - 10 = (x + 5)(x - 2)

$3)3a^{2}+7a-6=0\\\\D=7^{2} -4*3*(-6)=49+72=121=11^{2}\\\\a_{1}=\frac{-7-11}{6} =-3\\\\a_{2} =\frac{-7+11}{6}=\frac{2}{3}$

3a² + 7a - 6 = 3(a + 3)(a - 2/3)