given that √2 is a zero of 6x³+√2x²-10x-4√2 find it's other two zeros
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√2 is zero of 6x³ +√2x² -10x -4√2 , So, (x - √2) is a factor of 6x³ + √2x² -10x -4√2 .
6x³ + √2x² -10x -4√2 = 6x³ -6√2x² + 7√2x² -14x + 4x - 4√2= 6x²(x - √2) + 7√2x(x -2) + 4(x -√2) = {6x² + 7√2x + 4}(x - √2)= {6x² + 3√2x + 4√2x + 4}(x -√2)= {3√2x(√2x + 1) + 4(√2x +1)}(x -√2)= (3√2x +4)(√2x +1)(x - √2)
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6x³ + √2x² -10x -4√2 = 6x³ -6√2x² + 7√2x² -14x + 4x - 4√2= 6x²(x - √2) + 7√2x(x -2) + 4(x -√2) = {6x² + 7√2x + 4}(x - √2)= {6x² + 3√2x + 4√2x + 4}(x -√2)= {3√2x(√2x + 1) + 4(√2x +1)}(x -√2)= (3√2x +4)(√2x +1)(x - √2)
Read more on Brainly.in - https://brainly.in/question/1331555#readmore
indresh834:
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