=yz(x^2+5x-14)

=yz(x^2-2x+7x-14)

=yz[x(x-2)+7(x-2)

=yz(x-2)(x+7)

x⁸ + x⁴ + 1

= x⁸ + 2x⁴ + 1 - x⁴

= (x⁴ + 1)² - x⁴

= (x⁴ + 1)² - (x²)²

= (x⁴ + 1 + x²)(x⁴ + 1 - x²)

= (x⁴ + x² + 1)(x⁴ - x² + 1)

x³-3x²-4=

a. $6x^2-11x=x(6x-11)$
b. $x^7+x^5+1=(x^7-x)+(x^5-x^2)+x+x^2+1$

$=x(x^6-1)+x^2(x^3-1)+(x^2+x+1)$
$=x(x^3-1)(x^3+1)+x^2(x^3-1)+(x^2+x+1)$
$=(x^3-1)(x^4+x+x^2)+(x^2+x+1)$

$=(x-1)(x^2+x+1)(x^4+x^2+x)+(x^2+x+1)$
$=(x^2+x+1)[(x-1)(x^4+x^2+x)+1]$

$=(x^2+x+1)(x^5-x^4+x^3-x+1)$

c.

$x^8+x^4+1=(x^4)^2+2.x^4+1-x^4$

$=(x^4+1)^2-(x^2)^2$

$=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)(x^4+2x^2+1-x^2)$

$=(x^4-x^2+1)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

d.

$x^3-5x+8-4=x^3-5x+4$

$=x^3-x^2+x^2-x-(4x-4)$

$=x^2(x-1)+x(x-1)-4(x-1)=(x-1)(x^2+x-4)$

e.

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

$=x^2(x^3-1)+x(x^3-1)+x^2+x+1$

$=(x^3-1)(x^2+x)+(x^2+x+1)$
$=(x-1)(x^2+x+1)(x^2+x)+(x^2+x+1)$

$=(x^2+x+1)[(x-1)(x^2+x)+1]$

$=(x^2+x+1)(x^3-x+1)$

a: \(4x^2-x-5=\left(4x-5\right)\left(x+1\right)\)

b: \(x^2-2x-15=\left(x-5\right)\left(x+3\right)\)

a)x³+x²+4

=x³-x²+2x+2x²-2x+4

=x(x²-x+2)+2(x²-x+2)

=(x+2)(x²-x+2)

b)x³-2x-4

=x³+2x²+2x-2x²-4x-4

=x(x²+2x+2)-2(x²+2x+2)

=(x-2)(x²+2x+2)

\(=x^4+2x^2+1-\left(\sqrt{2}x\right)^2\)

\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)

\(=\left(x^2+1-\sqrt{2}x\right)\left(x^2+1+\sqrt{2}x\right)\)

\(x^4+1\)

\(=x^4+2x^2+1-2x^2\)

\(=\left(x^2+1\right)^2-\left(x\sqrt{2}\right)^2\)

\(=\left(x^2-x\sqrt{2}+1\right)\left(x^2+x\sqrt{2}+1\right)\)

bằng sau này bn nhá

x²-10x+16=x²-8x-2x+16=(x²-8x)-(2x-16)=x(x-8)-2(x-8)=(x-8)(x-2)

Đa thức này không phân tích được đâu bạn