1/ Phân tích đa thức thành nhân tử:
\(x^4+6x^3+7x^2-6x+1\)
2/ Tìm đa thức bậc ba P(x), biết P(x) chia cho x-1; x-2; x-3 đều dư 6 và P(-1)= -18.
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\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+6x^3+9x^2-2x^2-6x+1\)
\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2-2\left(x^2+3x\right).1+1^2\)
\(=\left(x^2+3x-1\right)^2\)
Chúc bạn học tốt.
\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+6x^3+9x^2-2x^2-6x+1\)
\(=x^2\left(x+3\right)^2-2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1=\left(x^2+3x-1\right)^2\)
x^4+6x^3+7x^2–6x+1
=x^4+(6x^3–2x^2)+(9x^2–6x+1)
= x^4+2x^2(3x–1)+(3x–1)^2
=(x^2+3x–1)^2
x4+6x3+7x2-6x+1
=(x4-2x2+1)+(6x3-6x)+9x2
=(x2-1)2+6x(x2-1)+9x2
=(x2-1).(x2-1+6x)+9x2
=(x2+3x-1)2
x4+6x3+7x2-6x+1
=(x4-2x2+1)+(6x3-6x)+9x2
=(x2-1)2+6x(x2-1)+9x2
=(x2-1). (x2-1+6x)+9x2
=(x2+3x-1)2
\(x^4-6x^3+7x^2-6x+1\)
\(=x^4+x^2+1-6x^3+6x^2-6x\)
\(=\left(x^2+1\right)^2-x^2-6x\left(x^2-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)-6x\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1-6x\right)\)
\(=\left(x^2-x+1\right)\left(x^2-5x+1\right)\)
= x4 - x3 + x2 - 5x3 + 5x2 - 5x + x2 - x +1 = x2 ( x2 - x +1 ) - 5x ( x2 - x +1 ) + x2 - x +1 = ( x2 - x +1 ) ( x2 - 5x + 1 )
a, x^2 + 5x +4
= x^2 + 1x + 4x + 4
= (x^2 + 1x) + (4x + 4)
= x ( x + 1 ) + 4 ( x + 1 )
= (x + 1) (x + 4)
b, x^2 - 6x + 5
= x^2 - 1x - 5x + 5
= (x^2 - 1x) - (5x - 5)
= x (x - 1) - 5 (x - 1)
= (x - 1) (x - 5)
c, x^2 + 7x + 12
= x^2 + 3x + 4x + 12
= (x^2 + 3x) + (4x + 12)
= x (x + 3) + 4 (x + 3)
= (x + 3) (x + 4)
d, 2x^2 - 5x + 3
= 2^x2 - 2x - 3x + 3
= 2x (x - 1) - 3 (x - 1)
= (x-1) (2x - 3)
e, 7x - 3x^2 - 4
= 3x + 4x - 3x^2 - 4
= (3x - 3x^2) + (4x - 4)
= 3x (1 - x) + 4 (x - 1)
= 3x (1-x) - 4 (1 - x)
= (1 - x) (3x - 4)
f, x^2 - 10x + 16
= x^2 - 2x - 8x + 16
= (x^2 - 2x) - (8x - 16)
= x (x - 2) - 8 (x - 2)
= (x - 2) (x - 8)
a, (x+1)(x+4)
b,(x-5)(x-1)
c,(x+3)(x+4)
d,(2x-3)(x-1)
e,(-3x+4)(x-1)
f, (x-8)(x-2)
Ta có: \(P\left(x\right)=x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=x^4+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)