Find the zeroes of the polynomial . ( Pls explain the solution )
Answers
Answer:
Given, f(x)=x²-√2x-12
→we have to find the values of x such that f(x)=0
→x²-√2x-12=0
→x²-3√2x+2√2x-12=0
→x(x-3√2)+2√2(x-3√2)=0
→(x+2√2)(x-3√2)=0
so, -2√2 and 3√2 are the zeros of the given polynomial.
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Step-by-step explanation:
x2" was replaced by "x^2".
Step by step solution :
STEP 1:
Trying to factor by splitting the middle term
1.1 Factoring x2-2x-17
The first term is, x2 its coefficient is 1 .
The middle term is, -2x its coefficient is -2 .
The last term, "the constant", is -17
Step-1 : Multiply the coefficient of the first term by the constant 1 • -17 = -17
Step-2 : Find two factors of -17 whose sum equals the coefficient of the middle term, which is -2 .
-17 + 1 = -16
-1 + 17 = 16
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step
1
:
x2 - 2x - 17 = 0
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