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If the graph of a polynomial intersects the x – axis at three points, then the number of zeroes =
If the graph of a polynomial intersects the xaxis at three points, then the number of zeroes are 3 because number of zeroes of the polynomial are the number of the coordinates of the points where its graph intersects the xaxis.
If ‘α’ and ‘β’ are the zeroes of a quadratic polynomial x^{2}+ 5x − 5, then
The sum and product of the zeroes of the polynomial x^{2}−6x+8 are respectively
Sum of the zeroes of the polynomial And Product of the zeroes of the polynomial
A polynomial of degree 3 is called a cubic polynomial.
For example, x^{3}−1, 4a^{3 }− 100a^{2 }+ a − 6, and m^{2}n + mn^{2} are all cubic polynomials with atmost 3 zeroes (having degree 3).
Hence, the polynomial having atmost 3 zero is a cubic polynomial.
Given: x^{3}+x^{2}−2x−3 = (x−2)(x^{2}+ax+b)+5
Dividing L.H.S. by (x−2)
∴ (x2)(x^{2}+3x+4)+5 = (x2)(x^{2}+az+b)+5
Comparing both side, we have a = 3, b = 4
The graph of the polynomial f(x) = 2x – 5 intersects the x – axis at
If The graph of the polynomial f(x) = 2x – 5 intersects the x – axis then y = 0
2x−5 = 0 ⇒ x = 5/2
If ‘α ’ and ‘β ’ are the zeroes of a quadratic polynomial x^{2}− 5x + b and α − β = 1, then the value of ‘b’ is
Here ……….(i)
And it is given that α−β = 1 ……….(ii)
On solving eq. (i) and eq. (ii),
we get
If one of the zeroes of the cubic polynomial x^{3}−7x+6 is 2, then the product of the other two zeroes is
Let α,β,γ are the zeroes of the given polynomial. Given: α = 2
Since αβγ = d/a
The maximum number of zeroes that a polynomial of degree 3 can have is
The maximum number of zeroes that a polynomial of degree 3 can have is three because the number of zeroes of a polynomical is equals to the degree of that polynomial.
The real number which should be subtracted from the polynomial 2x^{3}+5x^{2}−14x+10 so that the polynomial 2x−3 divides it exactly is
Therefore, 7 should be subtracted.
The graph of a cubic polynomial x^{3} – 4x meets the x – axis at (– 2, 0), (0, 0) and (2, 0), then the zeroes of the polynomial are
If the graph of a cubic polynomial intersects thrice xaxis, then the zeroes of the cubic polynomial are coordinates of xaxis.
If one zero of the quadratic polynomial x^{2}+ 3x + k is 2, then the value of ‘k’ is
According to question, p(2) = 0
⇒ (2)^{2}+3×2+k = 0
⇒ 4+6+k = 0
⇒ k = −10
If the sum of the zeroes of the cubic polynomial 4x^{3}−kx^{2}−8x−12 is 3/4 then the value of ‘k’ is
Let α,β,γ are the zeroes of the given polynomial. Given: α+β+γ = (3/4)
If α and β are the zeroes of the polynomial 2x^{2}+5x+1, then the value of α+β+αβ is
If √2 and −√2 are the zeroes of 2x^{4}−3x^{3}−3x^{2}+6x−2, then the other zeroes are
Since √2 and −√2 are the zeroes of 2x^{4}−3x^{3}−3x^{2}+6x−2, then(x−√2) (x+√2) are the factors of given polynomial i.e.,
(x−√2) (x+√2) = (x^{2}−2) is a factor of given polynomial.
∴p(x) = 2x^{4}−3x^{3}−3x^{2}+6x−2
⇒ p(x) = (x^{2}−2)(2x^{2}−3x+1)
⇒ p(x) = (x^{2}  2) [2x^{2}  2x  x + 1]
⇒ p (x) = (x^{2}  2) [2x (x 1)  1 (x  1)]
⇒ p (x) = (x^{2}  2) (x 1) (2x 1)
∴ Other zeroes are x  1 = 0 and 2x  1 = 0 ⇒ x = 1 and x = 1/2
If one zero of the polynomial p(x) = (k+4)x^{2}+13x+3k is reciprocal of the other, then the value of ‘k’ is
Let one zero of the given polynomial be athen the other zero be
⇒ k + 4 = 3k ⇒ 2k = 4 ⇒ k = 2
If two of the zeroes of a cubic polynomial ax^{3}+bx^{2}+cx+d are zero, then the third zero is
If ‘2’ is the zero of both the polynomials 3x^{2}+mx−14 and 2x^{3}+nx^{2}+x−2, then the value of m – 2n is
According to the question, p (2) = 3x^{2} + mx  14 = 0
⇒ 3(2)^{2} + m x 2  14 = 0
⇒ 12 + 2m  14 = 0 ⇒ m = 1
Also p(2) = 2x^{3 }+ nx^{2} + x  2 = 0
If one zero of the polynomial p(x) = (a^{2}+9)x^{2}+45x+6a is reciprocal of the other, then the value of ‘a’ is
Let the zeroes be k and (1/k)
Given quadratic polynomial is p(x) = (a^{2} + 9)x^{2} + 45x + 6a
Recall the sum of zeroes (k + 1/k) = –45/(a^{2} + 9)
Product of zeroes [k × (1/k)] = 6a/(a^{2} + 9)
⇒ 6a/(a^{2} + 9) = 1
⇒ (a^{2} + 9) = 6a
⇒ a^{2} – 6a + 9 = 0
⇒ a^{2} – 2(a)(3) + 3^{2} = 0
⇒ (a – 3)^{2} = 0
⇒ (a – 3) = 0
∴ a = 3
The polynomial to be added to the polynomial x^{4}+2x^{3}−2x^{2}+x−1 so that the resulting polynomial is exactly divisible by x^{2}+2x−3 is
Now (x^{2}+2x−3)−(x^{2}+x−1) = x−2
Therefore, (x−2) is the polynomial which to be added to the given polynomial.
The sum of two zeroes of the polynomial f(x) = 2x^{2 }+ (p+3)x + 5 is zero, then the value of ‘p’ is
Let one zeroes of the given polynomial be α and β. According to the question,
Sum of the zeroes = b/a = 0
⇒ (p3)/2 = 0
⇒  (p3) = 0×2
⇒  (p3) = 0
⇒ p =  3
A quadratic polynomial whose zeroes are  3 and 6, is
A quadratic polynomial whose product and sum of zeroes are 1/3 and √2 respectively is
Given:
On comparing, we get,a = 3,b = −3√2–,c = 1
Putting these values in general form of a quadratic polynomial ax^{2}+bx+c, we have 3x^{2}−3√2x+1
The sum and product of the zeroes of the polynomial f(x) = 4x^{2}−27x+3k^{2} are equal, then the value of ‘k’ is
Let α,β are the zeroes of the given polynomial.
Given:
⇒ −(−27) = 3k^{2 }⇒ k^{2} = 0 ⇒ k= ±3
A polynomial of degree ____ is called a quadratic polynomial.
The term quadratic describes something that pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms.It involves the second and no higher power of an unknown quantity or variable.
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