1078 Вправа 1078Алгебра 8 клас Бевз 2025 · § 21 · стор. 204

1078. Розклади на множники квадратний тричлен. а) x² − 10x + 21; б) a² + 2a − 15; в) −2x² − 5x + 3; г) c² − 11c − 26; ґ) 9a² + 24a + 16; д) 4c² + 25c + 25.

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

а) ${\displaystyle x^2-10x+21}$: ${\displaystyle x_1+x_2=}$${\displaystyle 10}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 21}$ → ${\displaystyle x_1=}$${\displaystyle 7,x_2=}$${\displaystyle 3}$;

${\displaystyle x^2-10x+21=}$${\displaystyle \left(x-7\right)\left(x-3\right)}$;

б) ${\displaystyle a^2+2a-15}$: ${\displaystyle a_1+a_2=}$${\displaystyle -2}$;

${\displaystyle a_1{a}_2=}$${\displaystyle -15}$ → ${\displaystyle a_1=}$${\displaystyle 3,a_2=}$${\displaystyle -5}$;

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

в) ${\displaystyle -2x^2-5x+3}$: ${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot (-2)\cdot 3=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm 7}{-4}}$;

${\displaystyle x_1=}$${\displaystyle 0,5}$;

${\displaystyle x_2=}$${\displaystyle -3}$;

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

г) ${\displaystyle c^2-11c-26}$: ${\displaystyle c_1+c_2=}$${\displaystyle 11}$;

${\displaystyle c_1{c}_2=}$${\displaystyle -26}$ → ${\displaystyle c_1=}$${\displaystyle 13,c_2=}$${\displaystyle -2}$;

${\displaystyle c^2-11c-26=}$${\displaystyle \left(c-13\right)\left(c+2\right)}$;

ґ) ${\displaystyle 9a^2+24a+16}$: ${\displaystyle D=}$${\displaystyle 24^2-4\cdot 9\cdot 16=}$${\displaystyle 0}$;

${\displaystyle a_0=}$${\displaystyle \frac{-24}{18}=}$${\displaystyle -\frac{4}{3}}$;

${\displaystyle 9a^2+24a+16=}$${\displaystyle 9{(a+\frac{4}{3})}^2=}$${\displaystyle {\left(3a+4\right)}^2}$;

д) ${\displaystyle 4c^2+25c+25}$: ${\displaystyle D=}$${\displaystyle 25^2-4\cdot 4\cdot 25=}$${\displaystyle 225}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{-25\pm 15}{8}}$;

${\displaystyle c_1=}$${\displaystyle -1,25}$;

${\displaystyle c_2=}$${\displaystyle -5}$;

${\displaystyle 4c^2+25c+25=}$${\displaystyle 4(c+1,25)(c+5)=}$${\displaystyle \left(4c+5\right)\left(c+5\right)}$.