phân tích thành nhân tử (đoán nghiệm ) \(2x^2-5x^2+8x-3\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,2x^3+3x^2-8x+3\)
\(=2x^3-2x^2+5x^2-5x-3x+3\)
\(=2x^2\left(x-1\right)+5x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x^2+5x-3\right)\left(x-1\right)\)
\(=\left(2x-1\right)\left(x+3\right)\left(x-1\right)\)
\(2,x^3-5x^2+2x+8\)
\(=x^3+x^2-6x^2-6x+8x+8\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x^2-6x+8\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\)
\(3,-6x^3+x^2+5x-2\)
\(=-6x^3-6x^2+7x^2+7x-2x-2\)
\(=-6x^2\left(x+1\right)+7x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(-6x^2+7x-2\right)\left(x+1\right)\)
\(=\left(-6x^2-3x-4x-2\right)\left(x+1\right)\)
\(=\left[-3x\left(2x+1\right)-2\left(2x+1\right)\right]\left(x+1\right)\)
\(=\left(-3x-2\right)\left(2x+1\right)\left(x+1\right)\)
\(4,3x^3+19x^2+4x-12\)
\(=3x^3+18x^2+x^2+6x-2x-12\)
\(=3x^2\left(x+6\right)+x\left(x+6\right)-2\left(x+6\right)\)
\(=\left(3x^2+x-2\right)\left(x+6\right)\)
\(=\left(3x-2\right)\left(x+1\right)\left(x+6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^3-5x^2+8x-3\)
\(\Leftrightarrow2x^3-x^2-4x^2+2x+6x+3\)
\(\Leftrightarrow x^2\cdot\left(2x-1\right)-2x\cdot\left(2x-1\right)+3\cdot\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-1\right)\cdot\left(x^2-2x+3\right)\)
\(2x^3-5x^2+8x-3\)
\(=2x^3-x^2-4x^2+2x+6x-3\)
\(=x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
= 2x^3 - 4x^2 - x^2 + 2x + 6x - 3
= 2x^2 ( x - 1/2 ) - x ( x - 1/2 ) +3 ( x - 1/2 )
= ( x - 1/2 )( 2x^2 - x + 3 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^3-5x^2+8x-3\)
\(=2x^3-x^2-4x^2+2x+6x-3\)
\(=x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nham nghiem thi ta thay pt co 1 ngiem x=2/3
=> tach nhu sau :
\(3x^3-2x^2-3x^2+2x+3x-2\)
\(3x^2\left(x-\frac{2}{3}\right)-3x\left(x-\frac{2}{3}\right)+3\left(x-\frac{2}{3}\right)\)
\(\left(x-\frac{2}{3}\right)\left(3x^2-3x+3\right)\)
Chuc ban hoc tot
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)
\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)
\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)
\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)
\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)
b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
Đè đúng là gì nè : \(2x^3-5x^2+8x-3\)
Đa thức có nghiệm là : \(\frac{1}{2}\) nên sẽ có một nhân tử là \(\left(2x-1\right)\)
\(2x^3-5x+8x-3\)
\(=2x^3-x^2-4x^2+2x+6x-3\)
\(=x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\).
Mình nghĩ là đề sai .
Đè đùng có thể là như thế này
\(2x^3-5x^2+8x-3\)
Dùng máy tính ta tính được nghiệm của đa thức là 0,5 nên khi phân tích ra nhân tử sẽ có \(\left(x-0,5\right)\)
Từ đó ta phân tích :
\(2x^3-5x^2+8x-3\)
\(=2x^2\left(x-0,5\right)-4x\left(x-0,5\right)+6\left(x-0,5\right)\)
\(=\left(x-0,5\right)\left(2x^2-4x+6\right)\)