Question

solve each polynomial equation by factoring

1. x^6-25x^4=0

2. y^3-3y^2=16y-48

3. x^3+2x^2-9x-18=0

4. x^3+x^2-4x-4=0

5. x^3-7x^2=-5x-35

6. a^4+2a^3+a^2==0

7. 3n^4-4n^2=-1

8. 8x^5+10x^4=4x^3+5x^2

9. 2n^4-9n^2+4=0

10. 8y^4-4y^2=0

Answer 1

Conditions

1. $x^{\wedge}6 - 25x^{\wedge}4 = 0$

2. $y^{\wedge}3 - 3y^{\wedge}2 = 16y - 48$

3. $x^{\wedge}3 + 2x^{\wedge}2 - 9x - 18 = 0$

4. $x^{\wedge}3 + x^{\wedge}2 - 4x - 4 = 0$

5. $x^{\wedge}3 - 7x^{\wedge}2 = -5x - 35$

6. $a^{\wedge}4 + 2a^{\wedge}3 + a^{\wedge}2 == 0$

7. $3n^{\wedge}4 - 4n^{\wedge}2 = -1$

8. $8x^{\wedge}5 + 10x^{\wedge}4 = 4x^{\wedge}3 + 5x^{\wedge}2$

9. $2n^{\wedge}4 - 9n^{\wedge}2 + 4 = 0$

10. $8y^{\wedge}4 - 4y^{\wedge}2 = 0$

Solution

1.

$$x ^ {6} - 2 5 x ^ {4} = 0$$ $$x ^ {4} (x ^ {2} - 2 5) = 0$$ $$x = 0, x = 5, x = - 5$$

2.

$$y ^ {3} - 3 y ^ {2} = 1 6 y - 4 8$$ $$y ^ {2} (y - 3) = 1 6 (y - 3)$$ $$y = 3, y = 4, y = - 4$$

3.

$$x ^ {3} + 2 x ^ {2} - 9 x - 1 8 = 0$$ $$x ^ {2} (x + 2) - 9 (x + 2) = 0$$ $$x = - 2, x = 3, x = - 3$$

4.

$x^{3}+x^{2}-4x-4=0$

$x^{2}(x+1)-4(x+1)=0$

$x=-1,x=2,x=-2$

5.

$x^{3}+7x^{2}=-5x-35$

$x^{2}(x+7)+5(x+7)=0$

$x=-7,x=\sqrt{5},x=-\sqrt{5}$

6.

$a^{4}+2a^{3}+a^{2}=0$

$a^{2}(a^{2}+2a+1)=a^{2}(a+1)^{2}=0$

$a=-1,a=0$

7.

$3n^{4}-4n^{2}=-1$

$3n^{4}-4n^{2}+1=0$

$D=16-12=4$

$n^{2}={\frac {4\pm 2}{6}},n^{2}=1,n^{2}={\frac {1}{3}}$

$n=\pm 1\,,n=\pm {\frac {1}{\sqrt {3}}}$

8.

$8x^{3}+10x^{4}=4x^{3}+5x^{2}$

$2x^{4}(4x+5)=x^{2}(4x+5)$

$x^{2}(2x^{2}-1)(4x+5)=0$

$x=-{\frac {5}{4}},x=0,x=\pm {\frac {1}{\sqrt {2}}}$

9.

$2n^{4}-9n^{2}+4=0$

$$\begin{array}{l} D = 81 - 32 = 49 \\ n = \frac{9 \pm 7}{4}, n = 4, n = \frac{1}{2} \\ n = \pm 2, n = \pm \frac{1}{\sqrt{2}} \\ \end{array}$$

10.

$$8y^4 - 4y^2 = 0$$ $$4y^2(2y^2 - 1) = 0$$ $$y = 0, y = \pm \frac{1}{\sqrt{2}}$$