812° Вправа 812Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 243

812. Розкладіть на лінійні множники квадратний тричлен:

1) ${\displaystyle x^2-3x-10=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-10)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

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

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

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

${\displaystyle x^2-3x-10=}$${\displaystyle \left(x+2\right)\left(x-5\right)}$;

2) ${\displaystyle x^2-10x+24=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-10)}^2-4\cdot 1\cdot 24=}$${\displaystyle 100-96=}$${\displaystyle 4}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{10\pm \sqrt{4}}{2\cdot 1}=}$${\displaystyle \frac{10\pm 2}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{10+2}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{10-2}{2}=}$${\displaystyle 4}$;

${\displaystyle x^2-10x+24=}$${\displaystyle \left(x-4\right)\left(x-6\right)}$;

3) ${\displaystyle -x^2+16x-15=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {16}^2-4\cdot (-1)\cdot (-15)=}$${\displaystyle 256-60=}$${\displaystyle 196}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-16\pm \sqrt{196}}{2\cdot (-1)}=}$${\displaystyle \frac{-16\pm 14}{-2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-16+14}{-2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-16-14}{-2}=}$${\displaystyle 15}$;

${\displaystyle -x^2+16x-15=}$${\displaystyle -\left(x-1\right)\left(x-15\right)}$;

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

${\displaystyle D=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-15)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{2\pm \sqrt{64}}{2\cdot 1}=}$${\displaystyle \frac{2\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{2+8}{2}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{2-8}{2}=}$${\displaystyle -3}$;

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

5) ${\displaystyle -x^2+4x-3=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 4^2-4\cdot (-1)\cdot (-3)=}$${\displaystyle 16-12=}$${\displaystyle 4}$;

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

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

${\displaystyle x_2=}$${\displaystyle \frac{-4-2}{-2}=}$${\displaystyle 3}$;

${\displaystyle -x^2+4x-3=}$${\displaystyle -\left(x-1\right)\left(x-3\right)}$;

6) ${\displaystyle x^2-6x-7=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot (-7)=}$${\displaystyle 36+28=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{6\pm \sqrt{64}}{2\cdot 1}=}$${\displaystyle \frac{6\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{6+8}{2}=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle \frac{6-8}{2}=}$${\displaystyle -1}$;

${\displaystyle x^2-6x-7=}$${\displaystyle \left(x+1\right)\left(x-7\right)}$;

7) ${\displaystyle x^2+4x+4=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 4^2-4\cdot 1\cdot 4=}$${\displaystyle 16-16=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle \frac{-4}{2}=}$${\displaystyle -2}$;

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

8) ${\displaystyle -x^2+10x-25=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {10}^2-4\cdot (-1)\cdot (-25)=}$${\displaystyle 100-100=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle \frac{-10}{-2}=}$${\displaystyle 5}$;

${\displaystyle -x^2+10x-25=}$${\displaystyle -{\left(x-5\right)}^2}$.