Question

don't scam
if u can't resist then
I can also not resist to report you:(^^)
or just solve 6x^2+7√2x+4
using splitting the middle term​

Answer 1

Answer:

I don't know ok

Step-by-step explanation:

next time apka kismat kharab hai

Answer 2

Answer:

• Polynomial = 6x³ + √2x² - 10x - 4√2
• One root = √2

If it's given that x = √2, then (x - √2) will certainly divide polynomial completely.

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

$\boxed{\begin{array}\quad\begin{tabular}{m{3.5em}cccc}&& 6x^2& + 7\sqrt{2}x& + 4\\\cline{1-6}\multicolumn{2}{l}{x - \sqrt{2}\big)}&6x^3& + \sqrt{2} x^2&- 10x& - 4\sqrt{2} \\&& 6x^3&- 6\sqrt{2} x^2&&\\\cline{3-5}&&&7\sqrt{2}x^2&- 10x&\\&&&7\sqrt{2}x^2& - 14x &\\\cline{4-5}&&&& 4x& - 4\sqrt{2} \\&&&&4x& - 4\sqrt{2} \\\cline{5-6}&&&&0&0\\\end{tabular}\end{array}}$

$\rule{200px}{.3ex}$

The quotient which we obtained contains two roots of the given polynomial, Now we will split it :

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$\dashrightarrow\sf 6x^2+7\sqrt{2}x+4 = 0\\\\\\\dashrightarrow\sf 6x^2 + (4\sqrt{2} + 3\sqrt{2})x + 4 = 0\\\\\\\dashrightarrow\sf 6x^2 + 4\sqrt{2}x + 3\sqrt{2}x + 4 = 0\\\\\\\dashrightarrow\sf 2x(3x + 2\sqrt{2}) + \sqrt{2}(3x + 2\sqrt{2}) = 0\\\\\\\dashrightarrow\sf (2x + \sqrt{2})(3x + 2\sqrt{2}) = 0\\\\\\\dashrightarrow\sf (2x + \sqrt{2}) = 0 \qquad(3x + 2\sqrt{2}) = 0\\\\\\\dashrightarrow\sf x = \dfrac{ - \sqrt{2}}{2} \qquad x = \dfrac{ - 2\sqrt{2}}{3}$

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