divide 2x⁴-13x³+19x²+7x-3 by x²
Answers
Step-by-step explanation:
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Answer:
The other two zeros are 3 and -\dfrac{1}{2}−
2
1
Step-by-step explanation:
Given: 2x^4-13x^3+19x^2+7x-32x
4
−13x
3
+19x
2
+7x−3
Two zeros are 2+\sqrt{3}2+
3
and 2-\sqrt{3}2−
3
The factor of the given polynomial,
(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))(x−(2+
3
))(x−(2−
3
))
x^2-4x+1x
2
−4x+1
Now divide the polynomial with their factor
\Rightarrow \dfrac{2x^4-13x^3+19x^2+7x-3}{x^2-4x+1}⇒
x
2
−4x+1
2x
4
−13x
3
+19x
2
+7x−3
\Rightarrow 2x^2-5x-3⇒2x
2
−5x−3
Now factor the above quadratic
\Rightarrow 2x^2-6x+x-3⇒2x
2
−6x+x−3
\Rightarrow 2x(x-3)+1(x-3)⇒2x(x−3)+1(x−3)
\Rightarrow (x-3)(2x+1)⇒(x−3)(2x+1)
The zeros are,
x - 3 = 0 and 2x + 1 = 0
x=3,x=-\dfrac{1}{2}x=3,x=−
2
1
Step-by-step explanation:
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