Phân tích thành nhân tử:
\(2x^2 -17x+19\)
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\(2x^3-12x^2+17x-2\)
\(=2x^3-4x^2-8x^2+16x+x-2\)
\(=2x^2\left(x-2\right)-8x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2-8x+1\right)\)
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: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)
3x3-7x2+17x-5
=3x3-x2-6x2+2x+15x-5
= x2.(3x-1)-2x.(3x-1)+5.(3x-1)
= (3x-1)(x2-2x+5)
Ta có : \(3x^3-7x^2+17x-5\)
\(=\left(3x^3-x^2\right)-\left(6x^2-2x\right)+\left(15x-5\right)\)
\(=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
Lời giải:
$2x^2-17x+19=2(x^2-8,5x+4,25^2)-\frac{137}{8}$
$=2(x-4,25)^2-\frac{137}{8}=2[(x-\frac{17}{4})^2-\frac{137}{16}]$
$=2(x-\frac{17+\sqrt{137}}{4})(x-\frac{17-\sqrt{137}}{4})$