815 Вправа 815Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 243

815. Розкладіть на лінійні множники квадратний тричлен:

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

${\displaystyle D=}$${\displaystyle 5^2-4\cdot 3\cdot (-2)=}$${\displaystyle 25+24=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm \sqrt{49}}{2\cdot 3}=}$${\displaystyle \frac{-5\pm 7}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-5+7}{6}=}$${\displaystyle \frac{2}{6}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-7}{6}=}$${\displaystyle -2}$;

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

2) ${\displaystyle -2x^2+x+15=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 1^2-4\cdot (-2)\cdot 15=}$${\displaystyle 1+120=}$${\displaystyle 121}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-1\pm \sqrt{121}}{2\cdot (-2)}=}$${\displaystyle \frac{-1\pm 11}{-4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+11}{-4}=}$${\displaystyle \frac{10}{-4}=}$${\displaystyle -2,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-11}{-4}=}$${\displaystyle 3}$;

${\displaystyle -2x^2+x+15=}$${\displaystyle -2(x+2,5)(x-3)=}$${\displaystyle \left(-2x-5\right)\left(x-3\right)}$;

3) ${\displaystyle 9x^2+6x+1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 6^2-4\cdot 9\cdot 1=}$${\displaystyle 36-36=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle \frac{-6}{18}=}$${\displaystyle -\frac{1}{3}}$;

${\displaystyle 9x^2+6x+1=}$${\displaystyle 9{(x+\frac{1}{3})}^2=}$${\displaystyle {\left(3x+1\right)}^2}$;

4) ${\displaystyle -5x^2+6x-1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 6^2-4\cdot (-5)\cdot (-1)=}$${\displaystyle 36-20=}$${\displaystyle 16}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-6\pm \sqrt{16}}{2\cdot (-5)}=}$${\displaystyle \frac{-6\pm 4}{-10}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-6+4}{-10}=}$${\displaystyle \frac{-2}{-10}=}$${\displaystyle 0,2}$;

${\displaystyle x_2=}$${\displaystyle \frac{-6-4}{-10}=}$${\displaystyle 1}$;

${\displaystyle -5x^2+6x-1=}$${\displaystyle -5(x-0,2)(x-1)=}$${\displaystyle \left(1-5x\right)\left(x-1\right)}$.