If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

1 Answer

Answer :

The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

Related questions

Description : If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

Description : If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Last Answer : Given p,q,r are in A.P. then q=2p+r​.....(1). Now px2+qx+r=0 will have real root then q2−4pr≥0. or, 4(p+r)2​−4pr≥0 or, p2+r2−14pr≥0 or, r2−14rp+49p2≥48p2 or, (r−7p)2≥(43​p)2 or, (pr​−7)2≥(43​)2 [ Since p=0 for the given equation to be quadratic] or, ∣∣∣∣∣​pr​−7∣∣∣∣∣​≥43​.

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

Description : When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th

Last Answer : answer:

Description : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

Last Answer : answer:

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : As the given polynomial divisible by x-2 means the polynomial satisfies for the value x=2 So putting x=2 in x²+(4-k)x+2 yields 0 ⇒2²+(4-k)2+2=0 ⇒4+8-2k+2=0 ⇒ 2k=14 ⇒ k= ... ;-3x+2 if factorized yields (x-1)(x-2). Thus is divisible by x-2 as well as divisible by x-1.

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : The value of 'k' is 4

Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

Description : If the difference in the roots of the equation x^2 – px + q = 0 is unity, then which one of the following is correct ? -Maths 9th

Last Answer : answer:

Description : If x^2 + mx + n = 0 and x^2 + px + q = 0 have a common root, then the common root is -Maths 9th

Last Answer : Let α be the common root ∴α2+pα+q=0 ...........(1) and α2+qα+p=0 ........ (2) Solving (1) & (2), we get, p2−q2α2​=q−pα​=q−p1​∴α=q−pp2−q2​ and α=1 ⇒q−pp2−q2​=1 ⇒p2−q2=q−p (or) (p2−q2)+(p−q)=0 ⇒(p−q)[p+q+1]=0 ⇒p−q=0 or p+q+1=0

Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

Last Answer : answer:

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

Description : Find the value of the polynomial p(x) = x^3-3x^2-2x+6 at x = underroot 2 -Maths 9th

Last Answer : In this chapter, we shall proceed with recalling some of the constructions already learnt in the earlier classes and deal with some more. Here in this section, we will construct some of these ... be done? 2. Always explain the construction. Write the sequence of steps that are actually taken.

Description : If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Last Answer : answer:

Description : Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th

Last Answer : Solution :-

Description : If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then: -Maths 9th

Last Answer : answer:

Description : Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using derivatives.

Last Answer : Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using ... `pqsqrtr` B. `2pqsqrtr` C. `2rsqrtpq` D. None of these

Description : If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. -Maths 9th

Last Answer : Here a = 3, b = -k, c = 6 Sum of the zeroes, (α + β) = − = 3 …..(given) ⇒ −(−)3 = 3 ⇒ k = 9

Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.

Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.

Description : If x+1 is a factor of the polynomial 3x(square) - kx,then find the value of k. -Maths 9th

Last Answer : Solution :-

Description : A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals : -Maths 9th

Last Answer : answer:

Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

Last Answer : (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

Last Answer : (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

Description : Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

Description : Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0

Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

Description : Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

Description : Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0

Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

Description : Check whether polynomial p(x) = 2x(cube) - 9x(square) + x + 12 is a multiple of 2x-3 or not. -Maths 9th

Last Answer : Solution :-

Description : write the degree of the polynomial p(x)=4 -Maths 9th

Last Answer : f(x)=4 =4*x0 So degree =0

Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

Description : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

Description : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

Description : If the polynomial x^19 + x^17 + x^13 + x^11 + x^7 + x^5 + x^3 is divided by (x^2 + 1), then the remainder is : -Maths 9th

Last Answer : answer:

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Description : Find the value of polynomial 12x(square) - 7x + 1, when x=1/4. -Maths 9th

Last Answer : Solution :-

Description : For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

Last Answer : f(x) = 3x3 + mx2 + 4x – 4m f(x) is divisible by (x + 2) if f(–2) = 0 Now f(–2) = 3(–2)3 + m(–2)2 + 4(–2) – 4m = – 24 + 4m – ... ; 4m = – 32 ≠ 0 ∴ No such value of m exists for which (x + 2) is a factor of the given expression

Description : For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th

Last Answer : Given f(x) = 3x³ - kx² + 4x + 16. Since (x - k/2) is a factor of polynomial. This means x = k/2 is the zero of the given polynomial. ⇒ f(k/2) = 3(k/2)³ - k(k/2)² + 4(k/2) + 16 ⇒ 0 ... - 4k + 32) + 4(k² - 4k + 32) ⇒ 0 = (k + 4)(k² - 4k + 32) ⇒ k = -4.

Description : Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th

Last Answer : Let the polynomial be f(x) = 5x – 4x2 + 3 Now, for x = 2, f(2) = 5(2) – 4(2)2 + 3 => f(2) = 10 – 16 + 3 = –3 Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 ... (–1) = –5 –4 + 3 = -6 The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

Description : If the remainder of the polynomial a0 + a1x + a2x^2 + ....... + anx^n when divided by (x – 1) is 1, then which one of the following is correct ? -Maths 9th

Last Answer : answer:

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

Description : When a polynomial f(x) is divided by (x – 3) and (x + 6), the respective remainders are 7 and 22. What is the remainder when f(x) is -Maths 9th

Last Answer : answer:

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.