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3: \(15x^3+29x^2-8x-12\)
\(=15x^3+30x^2-x^2-2x-6x-12\)
\(=\left(x+2\right)\left(15x^2-x-6\right)\)
\(=\left(x+2\right)\left(15x^2-10x+9x-6\right)\)
\(=\left(x+2\right)\left(3x-2\right)\left(3x+5\right)\)
5: \(x^3+9x^2+26x+24\)
\(=x^3+4x^2+5x^2+20x+6x+24\)
\(=\left(x+4\right)\left(x^2+5x+6\right)\)
\(=\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0
=> x3 + 4x2 - 29x + 24 = x2(x-1) + 5x(x-1) - 24(x-1)
= (x-1)(x2+5x-24) = (x-1)(x-3)(x+8)
b) ...
a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a) \(x^2-6x+8\)
\(=x^2-2\cdot x\cdot3+3^2-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
Còn lại tương tự
a) \(x^2-6x+8=x^2-2x-4x+8\)
\(=\left(x^2-2x\right)-\left(4x-8\right)\)
=x(x-2)-4(x-2) = (x-2)(x-4)
b mk thấy nó sai đề sao ý
c) \(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)
\(=\left(x^2+x+4\right)^2+2.4x.\left(x^2+x+4\right)+16x^2-x^2\)
\(=\left(x^2+x+4+4x\right)^2-x^2\)
\(=\left(x^2+5x+4\right)^2-x^2\)
\(=\left(x^2+5x+4-x\right)\left(x^2+5x+4+x\right)=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)
a, \(x^3+4x^2-29x+24\)
\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left[x\left(x-3\right)+8\left(x-3\right)\right]\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
\(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
Chúc bạn học tốt.
b, \(\left(x^2+x\right)^2+4x^2+4x-12=x^4+2x^3+x^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x^3+3x^2+8x+12\right)\left(x-1\right)\)
\(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\left(x-1\right)\)
\(=\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
c, \(x^3+3x^2-4=\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(2x+4\right)\)
\(=x^2\left(x+2\right)+x\left(x+2\right)-2\left(x+2\right)\)
= \(\left(x^2+x-2\right)\left(x+2\right)\)
a)\(x^5+x^4+1=x^5-\left(-x^3+x^3\right)+x^4+\left(x^2-x^2\right)+\left(x-x\right)+1\)
\(=x^5-x^3+x^2+x^4-x^2+x+x^3-x+1\)
\(=x^2\left(x^3-x+1\right)+x\left(x^3-x+1\right)+\left(x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
b,c có ng lm rồi
d)\(2x^4-3x^3-7x^2+6x+8\)
Ta thấy x=-1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x+1
\(\Rightarrow\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)
\(\Rightarrow\left(x+1\right)\left[2x^3-x^2-4x-4x^2+2x+8\right]\)
\(\Rightarrow\left(x+1\right)\left[x\left(2x^2-x-4\right)-2\left(2x^2-x-4\right)\right]\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
phần còn lại bạn tự lo nhé
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
a) \(x^{12}-3x^6+1\)
\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)
\(=\left(x^6-1\right)^2-\left(x^3\right)^2\)
\(=\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
1)
\(15x^3+29x^2-8x-12=(15x^3+30x^2)-(x^2+2x)-(6x+12)\)
\(=15x^2(x+2)-x(x+2)-6(x+2)\)
\(=(x+2)(15x^2-x-6)=(x+2)(15x^2-10x+9x-6)\)
\(=(x+2)[5x(3x-2)+3(3x-2)]\)
\(=(x+2)(3x-2)(5x+3)\)
2)
\(x^3+4x^2-29x+24=(x^3-x^2)+(5x^2-5x)-(24x-24)\)
\(=x^2(x-1)+5x(x-1)-24(x-1)\)
\(=(x-1)(x^2+5x-24)\)
\(=(x-1)(x^2-3x+8x-24)\)
\(=(x-1)[x(x-3)+8(x-3)]=(x-1)(x-3)(x+8)\)
3.
\(x^4+6x^3+7x^2-6x+1\)
\(=(x^4+6x^3+9x^2)-2x^2-6x+1\)
\(=(x^2+3x)^2-2x^2-6x+1\)
\(=(x^2+3x)^2-2(x^2+3x)+1=(x^2+3x-1)^2\)
4.
\(x^3+4x^2-31x-70=(x^3+2x^2)+(2x^2+4x)-(35x+70)\)
\(=x^2(x+2)+2x(x+2)-35(x+2)\)
\(=(x+2)(x^2+2x-35)\)
\(=(x+2)(x^2-5x+7x-35)\)
\(=(x+2)[x(x-5)+7(x-5)]\)
\(=(x+2)(x-5)(x+7)\)
5.
\(5x^4+24x^3-15x^2-118x+4\)
\(=5x^4-10x^3+34x^3-68x^2+53x^2-106x-(12x-24)\)
\(=5x^3(x-2)+34x^2(x-2)+53x(x-2)-12(x-2)\)
\(=(5x^3+34x^2+53x-12)(x-2)\)
\(=(5x^3+15x^2+19x^2+57x-4x-12)(x-2)\)
\(=[5x^2(x+3)+19x(x+3)-4(x+3)](x-2)\)
\(=(5x^2+19x-4)(x+3)(x-2)\)
\(=(5x^2+20x-x-4)(x+3)(x-2)\)
\(=[5x(x+4)-(x+4)](x+3)(x-2)\)
\(=(5x-1)(x+4)(x+3)(x-2)\)
6.
\(x^4-6x^3+7x^2+6x-8\)
\(=(x^4-6x^3+9x^2)-2x^2+6x-8\)
\(=(x^2-3x)^2-2(x^2-3x)-8\)
\(=(x^2-3x)^2+2(x^2-3x)-4(x^2-3x)-8\)
\(=(x^2-3x)(x^2-3x+2)-4(x^2-3x+2)\)
\(=(x^2-3x+2)(x^2-3x-4)\)
\(=(x^2-x-2x+2)(x^2+x-4x-4)\)
\(=[x(x-1)-2(x-1)][x(x+1)-4(x+1)]\)
\(=(x-1)(x-2)(x+1)(x-4)\)
Riêng đối với câu 5, cách làm dưới của mình hiệu quả chỉ khi bạn tìm được nghiệm của đa thức rồi tách ghép hợp lý. Nếu không, bạn có thể sử dụng pp hệ số bất định.
Đặt \(5x^4+24x^3-15x^2-118x+24=(5x^2+ax+b)(x^2+cx+d)\) (tất nhiên $a,b,c,d$ nguyên)
\(\Leftrightarrow 5x^4+24x^3-15x^2-118x+24=5x^4+5cx^3+5dx^2+ax^3+acx^2+axd+bx^2+bcx+bd\)
\(\Leftrightarrow 5x^4+24x^3-15x^2-118x+24=5x^4+x^3(5c+a)+x^2(5d+ac+b)+x(ad+bc)+bd\)
Đồng nhất hệ số:
\(\left\{\begin{matrix} 5c+a=24\\ 5d+ac+b=-15\\ ad+bc=-118\\ bd=24\end{matrix}\right.\)
Từ \(bd=24\) ta chọn $b=-4$, $d=-6$ (có nhiều TH, nhưng ta chọn một con số khả thi, nếu không được thì chuyển sang $-8,-3....$
\(\left\{\begin{matrix} 5c+a=24\\ ac=19\\ 6a+4c=118\end{matrix}\right.\Rightarrow a=19;c=1\)
Vậy \(5x^4+24x^3-15x^2-118x+24=(5x^2+19x-4)(x^2+x-6)\)
\(=(5x^2+20x-x-4)(x^2-2x+3x-6)\)
\(=[5x(x+4)-(x+4)][x(x-2)+3(x-2)]\)
\(=(x+4)(5x-1)(x-2)(x+3)\)