(1)$(2-\mathrm{\square }{)}^2=\mathrm{△}-◯+4b^{16}\backslash \; (2-\backslash square)^2=\backslash triangle-\backslash bigcirc+4b^\{16\}$

Wende die 2. Binomische Formel an:

$(a-b{)}^2=a^2-2ab+b^2\backslash hspace\{14mm\}(a-b)^2\backslash hspace\{4mm\}\; =\backslash hspace\{2mm\}a^2-2ab+b^2$

${(2-\boxed{{\displaystyle 2b^8}})}^2=\boxed{{\displaystyle 4}}-\boxed{{\displaystyle 8b^8}}+4b^{16}\backslash hspace\{12mm\}\backslash left(2-\backslash boxed\{2b^8\}\backslash right)^2=\backslash boxed\{\backslash \; 4\backslash \; \}-\backslash boxed\{8b^8\}\backslash \; +4b^\{16\}$

(2)$({\displaystyle \frac{1}{5}}{c}^3+\mathrm{\square })\cdot ({\displaystyle \frac{1}{5}}{c}^3-\mathrm{\square })\backslash \; \backslash left(\backslash dfrac\{1\}\{5\}c^3+\backslash square\backslash right)\backslash cdot\; \backslash left(\backslash dfrac\{1\}\{5\}c^3-\backslash square\backslash right)$$=◯-{\displaystyle \frac{4}{81}}{d}^4\backslash \; =\backslash \; \backslash bigcirc-\backslash dfrac\{4\}\{81\}d^4$

Wende die 3. Binomische Formel an:

$(a+b)\cdot (a-b)=a^2-b^2\backslash hspace\{8mm\}(a+b)\backslash hspace\{4mm\}\backslash hspace\{1mm\}\backslash cdot\backslash hspace\{5mm\}(a-b)\backslash hspace\{8mm\}=\backslash hspace\{7mm\}\; a^2-b^2$

$({\displaystyle \frac{1}{5}}{c}^3+\boxed{{\displaystyle \frac{2}{9}{d}^2}})\cdot ({\displaystyle \frac{1}{5}}{c}^3-\boxed{{\displaystyle \frac{2}{9}{d}^2}})=\boxed{{\displaystyle \frac{1}{25}{c}^6}}-{\displaystyle \frac{4}{81}}{d}^4\backslash \; \backslash left(\backslash dfrac\{1\}\{5\}c^3+\backslash boxed\{\backslash dfrac\{2\}\{9\}d^2\}\backslash right)\backslash cdot\; \backslash left(\backslash dfrac\{1\}\{5\}c^3-\backslash boxed\{\backslash dfrac\{2\}\{9\}d^2\}\backslash right)\backslash \; =\backslash \; \backslash boxed\{\backslash dfrac\{1\}\{25\}c^6\}-\backslash dfrac\{4\}\{81\}d^4$

$(x+\mathrm{\square })\cdot (x-\mathrm{△})=0(x+\backslash square)\backslash cdot\; (x-\backslash triangle)=0$

Die quadratische Gleichung hat zwei Nullstellen nämlich: $x_1=-4;x_2=3\backslash \; x\_\{1\}\backslash \; =\backslash \; -4;\backslash quad\; x\_\{2\}=3$

Linearfaktorenzerlegung:

$(x-x_1)\cdot (x-x_2)\backslash hspace\{40mm\}(x-x\_\{1\})\backslash cdot(x-x\_\{2\})\backslash qquad$setze$x_1\backslash \; x\_\{1\}\backslash$ und$x_2\backslash \; x\_\{2\}\backslash$ ein.

$(x+4)\cdot (x-3)=0\backslash hspace\{42mm\}(x+4)\backslash cdot(x-3)\backslash \; =0$

Ausmultipliziert ergibt das:

$x^2+x-12=0\backslash hspace\{49mm\}x^2+x-12\backslash \; =\backslash \; 0$