1381 Вправа 1381Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1381. Розклади на множники тричлен. а) x² − 7x + 10; б) x² − 9x + 18; в) y² − 2y − 35; г) y² − 4y − 60; ґ) a² − a − 56; д) c² − 5c − 24.

За теоремою Вієта знаходимо корені, тоді ${\displaystyle x^2+px+q=}$${\displaystyle (x-x_1)(x-x_2)}$.

а) ${\displaystyle x^2-7x+10}$: ${\displaystyle x_1+x_2=}$${\displaystyle 7}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 10}$ → ${\displaystyle x_1=}$${\displaystyle 5,x_2=}$${\displaystyle 2}$;

${\displaystyle x^2-7x+10=}$${\displaystyle \left(x-5\right)\left(x-2\right)}$;

б) ${\displaystyle x^2-9x+18}$: ${\displaystyle x_1+x_2=}$${\displaystyle 9}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 18}$ → ${\displaystyle x_1=}$${\displaystyle 6,x_2=}$${\displaystyle 3}$;

${\displaystyle x^2-9x+18=}$${\displaystyle \left(x-6\right)\left(x-3\right)}$;

в) ${\displaystyle y^2-2y-35}$: ${\displaystyle y_1+y_2=}$${\displaystyle 2}$;

${\displaystyle y_1{y}_2=}$${\displaystyle -35}$ → ${\displaystyle y_1=}$${\displaystyle 7,y_2=}$${\displaystyle -5}$;

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

г) ${\displaystyle y^2-4y-60}$: ${\displaystyle y_1+y_2=}$${\displaystyle 4}$;

${\displaystyle y_1{y}_2=}$${\displaystyle -60}$ → ${\displaystyle y_1=}$${\displaystyle 10,y_2=}$${\displaystyle -6}$;

${\displaystyle y^2-4y-60=}$${\displaystyle \left(y-10\right)\left(y+6\right)}$;

ґ) ${\displaystyle a^2-a-56}$: ${\displaystyle a_1+a_2=}$${\displaystyle 1}$;

${\displaystyle a_1{a}_2=}$${\displaystyle -56}$ → ${\displaystyle a_1=}$${\displaystyle 8,a_2=}$${\displaystyle -7}$;

${\displaystyle a^2-a-56=}$${\displaystyle \left(a-8\right)\left(a+7\right)}$;

д) ${\displaystyle c^2-5c-24}$: ${\displaystyle c_1+c_2=}$${\displaystyle 5}$;

${\displaystyle c_1{c}_2=}$${\displaystyle -24}$ → ${\displaystyle c_1=}$${\displaystyle 8,c_2=}$${\displaystyle -3}$;

${\displaystyle c^2-5c-24=}$${\displaystyle \left(c-8\right)\left(c+3\right)}$.