Advertisements
Advertisements
Question
Show that (x + 1) is a factor of the polynomial `x^3 - x^2 - (2 + sqrt(2))x + sqrt(2)`.
Solution
Let p(x) = `x^3 - x^2 - (2 + sqrt(2))x + sqrt(2)`
By factor theorem, (x + 1) will be the factor of p(x) if p(– 1) = 0.
Now, p(– 1) = `(-1)^3 - (-1)^2 - (2 + sqrt(2)) (-1) + sqrt(2)`
= `-1 - 1 + 2 + sqrt(2) + sqrt(2)`
= `2sqrt(2)`
∵ p(– 1) ≠ 0
Hence, (x + 1) is not a factor of p(x).
APPEARS IN
RELATED QUESTIONS
Solve : 7y = -3y2 - 4
If the roots of x² + kx + k = 0 are real and equal, what is the value of k?
If α and β are the roots of the quadratice equation x²- 2x - 7= 0, find the
value α² + β²
If one root of the quadratic, x2 - 7x + k = 0 is 4. then find the value of k.
Convert the following equations into simultaneous equations and solve:
`sqrt("x"/"y") = 4, 1/"x" + 1/"y" = 1/"xy"`
Choose the correct alternative answer for the following sub-questions and write the correct alphabet.
Which of the following quadratic equation has roots – 3 and – 5?
Write the roots of following quadratic equation.
(p – 5) (p + 3) = 0
If one of the roots of quadratic equation x2 + kx + 54 = 0 is – 6, then complete the following activity to find the value of ‘k’.
Activity: One of the roots of the quadratic equation x2 + kx + 54 = 0 is – 6.
Therefore let’s take x = ______
(– 6)2 + k(– 6) + 54 = 0
(______) – 6k + 54 = 0
– 6k + ______ = 0
k = ______
Roots of a quadratic equation are 5 and – 4, then form the quadratic equation
Solve the following quadratic equation.
`sqrt(3) x^2 + sqrt(2)x - 2sqrt(3)` = 0
Solve the following quadratic equations by formula method.
`y^2 + 1/3y` = 2
Form a quadratic equation if the roots of the quadratic equation are `2 + sqrt(7)` and `2 - sqrt(7)`
Solve the following quadratic equation.
`1/(4 - "p") - 1/(2 + "p") = 1/4`
If x = `sqrt(7) - 2`, find the value of `(x + 1/x)`.
Is (x – 5) a factor of the polynomial x3 – 5x – 30?
One of the roots of equation x2 + 5x + a = 0 is – 3. To find the value of a, fill in the boxes.
Since, `square` is a root of equation x2 + 5x + a = 0
∴ Put x = `square` in the equation
⇒ `square^2 + 5 xx square + a` = 0
⇒ `square + square + a` = 0
⇒ `square + a` = 0
⇒ a = `square`
Find the roots of the quadratic equation `x^2 - (sqrt(3) + 1)x + sqrt(3)` = 0.
If 3 is one of the roots of the quadratic equation kx2 − 7x + 12 = 0, then k = ______.