X=5
7(5)-8=35-8=27
3(5)+12=15+12=27
<h3>Given</h3>
tan(x)²·sin(x) = tan(x)²
<h3>Find</h3>
x on the interval [0, 2π)
<h3>Solution</h3>
Subtract the right side and factor. Then make use of the zero-product rule.
... tan(x)²·sin(x) -tan(x)² = 0
... tan(x)²·(sin(x) -1) = 0
This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:
... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)
Then our equation becomes
... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0
... -sin(x)²/(1 +sin(x)) = 0
Now, we know the only solutions are found where sin(x) = 0, at ...
... x ∈ {0, π}
If x=-5 is a zero, then the first factor of the polynomial would be (x + 5 )
To find the other two factors we can divide the polynomial by the expression (x+5).
Using synthetic division, we have:
-5 I 4 15 -24 5 (Coefficients of the dividend)
I -20 25 -5 (Multiplying each coefficient by the results of the substraction and adding)
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4 -5 1 0 (Coefficients of the quotient)
The result of the division is 4x^2 - 5x + 1. Factoring it, we have:
4x^2 - 4x -x + 1 (Separating -5x into -x and -4x)
4x (x - 1) - (x -1) (Factoring each pair of terms)
(x-1)(4x-1) (Factoring using the common factor)
So the answer would be:
(x + 5 )(x-1)(4x-1)
If I’m correct, then the answer is D. Someone please correct me if I’m wrong.