This is a cube root, so we look for factors of 162 which are perfect cubes.
Find the prime factors of 162:-
162 = 2 * 3 * 3 * 3* 3
27 = 3^3 is a perfect cube
162 = 6 * 27
so ^3√ 162 = ^3√6 * ^3√27 = ^3√6 * 3
so the simplest form is 3 ^3√6
1 1/16 is it's simplest form.
Answer:
3
Step-by-step explanation:
f(x) = 3x² - 2x + 1
f(x) = ax² + bx + c
a = 3
If I am reading the question correctly it would come out to -6x+42 First do all the addition and squaring, then factor in that negative. You come out to -2x^3 +2x^3-4x-2x+25+5+9+3. The x^3's cancel, the -x's add to -6x, then all that other adds to 42.
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