42 Вправа 42 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

42. Розкладіть квадратний тричлен на лінійні множники:

Використовуємо формулу ${\displaystyle a{x}^2+bx+c=}$${\displaystyle a(x-x_1)(x-x_2)}$.

1) ${\displaystyle x^2-7x+12=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-7)}^2-4\cdot 1\cdot 12=}$${\displaystyle 49-48=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{7\pm \sqrt{1}}{2\cdot 1}=}$${\displaystyle \frac{7\pm 1}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{7+1}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{7-1}{2}=}$${\displaystyle 3}$;

${\displaystyle x^2-7x+12=}$${\displaystyle \left(x-3\right)\left(x-4\right)}$;

2) ${\displaystyle x^2+2x-35=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle -2}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -35}$;

${\displaystyle x_1=}$${\displaystyle -7}$; ${\displaystyle x_2=}$${\displaystyle 5}$;

${\displaystyle x^2+2x-35=}$${\displaystyle \left(x+7\right)\left(x-5\right)}$;

3) ${\displaystyle 3x^2+5x+2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 5^2-4\cdot 3\cdot 2=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm \sqrt{1}}{2\cdot 3}=}$${\displaystyle \frac{-5\pm 1}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-5+1}{6}=}$${\displaystyle \frac{-4}{6}=}$${\displaystyle -\frac{2}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-1}{6}=}$${\displaystyle -1}$;

${\displaystyle 3x^2+5x+2=}$${\displaystyle 3(x+1)(x+\frac{2}{3})=}$${\displaystyle (x+1)\cdot 3(x+\frac{2}{3})=}$${\displaystyle \left(x+1\right)\left(3x+2\right)}$;

4) ${\displaystyle -6x^2+17x-5=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {17}^2-4\cdot (-6)\cdot (-5)=}$${\displaystyle 289-120=}$${\displaystyle 169}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-17\pm \sqrt{169}}{2\cdot (-6)}=}$${\displaystyle \frac{-17\pm 13}{-12}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-17+13}{-12}=}$${\displaystyle \frac{-4}{-12}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-17-13}{-12}=}$${\displaystyle \frac{-30}{-12}=}$${\displaystyle 2,5}$;

${\displaystyle -6x^2+17x-5=}$${\displaystyle -6(x-2,5)(x-\frac{1}{3})=}$${\displaystyle -2(x-2,5)\cdot 3(x-\frac{1}{3})=}$${\displaystyle \left(5-2x\right)\left(3x-1\right)}$.