{"id":25300,"date":"2020-03-07T15:25:38","date_gmt":"2020-03-07T09:55:38","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25300"},"modified":"2021-07-02T15:21:18","modified_gmt":"2021-07-02T09:51:18","slug":"kseeb-solutions-for-class-10-maths-chapter-9-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-9-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 9 Polynomials Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Additional Questions<\/h2>\n

I. Multiple Choice Questions:<\/span><\/p>\n

Question 1.
\nThe degree of polynomial is x + 2
\na. 2
\nb. 1
\nc. 3
\nd. 4
\nAnswer:
\nb. 1<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nThe degree of a quadratic polynomial is
\na. 0
\nb. 1
\nc. 2
\nd. 3
\nAnswer:
\nc. 2<\/p>\n

Question 3.
\nThe degree polynomial of a bi-quadratic
\na. 0
\nb. 2
\nc. 4
\nd. 1
\nAnswer:
\nc. 4<\/p>\n

Question 4.
\nThe standard polynomial. form of a linear
\na. ax + c
\nb. ax2<\/sup> + c
\nc. ax2<\/sup> + bx + c = 0
\nd. ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nAnswer:
\na. ax + c<\/p>\n

Question 5.
\nThe standard form of a quadratic equation.
\na. ax2<\/sup> + bx + c = 0
\nb. ax + c = 0
\nc. ax2<\/sup> + c = 0
\nd. ax3<\/sup> + bx + c = 0
\nAnswer:
\na. ax2<\/sup> + bx + c = 0<\/p>\n

Question 6.
\nA polynomial of degree 3 is called
\na. a linear polynomial
\nb. a quadratic polynomial
\nc. a cubic polynomial
\nd. a biquadratic polynomial
\nAnswer:
\nc. a cubic polynomial<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nThe value of px. = x2<\/sup> – 3x – 4, x = -1
\na. 1
\nb. – 4
\nc. 0
\nd. – 3
\nAnswer:
\nc. 0<\/p>\n

Question 8.
\nThe degree of the polynomial x4<\/sup> + x3<\/sup> is
\na. 2
\nb. 3
\nc. 5
\nd. 4
\nAnswer:
\nd. 4<\/p>\n

Question 9.
\nThe number of zeroes of linear polynomial at most is
\na. 0
\nb. 1
\nc. 2
\nd. 3
\nAnswer:
\nb. 1<\/p>\n

Question 10.
\nThe degree of polynomial x + 2. x + 1. is
\na. 1
\nb. 3
\nc. 4
\nd. 2
\nAnswer:
\nd. 2<\/p>\n

Question 11.
\nThe degree of a zero polynomial is
\na. not defined
\nb. 1
\nc. 2
\nd. 3
\nAnswer:
\na. not defined<\/p>\n

Question 12.
\nThe zeroes of the polynomial x3<\/sup> -4x are
\na. 0, \u00b1 2
\nb. 0, \u00b1 1
\nc. 0, \u00b1 3
\nd. 0, 0
\nAnswer:
\na. 0, \u00b1 2<\/p>\n

Question 13.
\nThe zeroes of the polynomials, t2<\/sup> – 15 are.
\na. \u00b1 \\(\\sqrt{15}\\)
\nb. \u00b1 \\(\\sqrt{5}\\)
\nc. \u00b1 \\(\\sqrt{3}\\)
\nd. \u00b1 3
\nAnswer:
\na. \u00b1 \\(\\sqrt{15}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nDividend is equal to.
\na. divisor \u00d7 quotient + remainder
\nb. divisor \u00d7 quotient
\nc. divisor \u00d7 quotient – remainder
\nd. divisor \u00d7 quotient \u00d7 remainder
\nAnswer:
\na. divisor \u00d7 quotient + remainder<\/p>\n

Question 15.
\nIf the divisor is x2<\/sup> and quotient is x while the number remainder is 1, then the dividend is.
\na. x2<\/sup>
\nb. x
\nc. x3<\/sup>
\nd. x3<\/sup> + 1
\nAnswer:
\nd. x3<\/sup> + 1<\/p>\n

Question 16.
\nWhat is the co-efficient of the first term of the quotient when 2x2<\/sup> + 2x + 1 is divided by x + 2?
\na. 1
\nb. 2
\nc. 3
\nd. – 2
\nAnswer:
\nb. 2<\/p>\n

Question 17.
\nThe zeroes of the polynomial x2<\/sup> – 3x – 4 are
\na. 4, – 1
\nb. 4, 1
\nc. – 4, 1
\nd.- 4, – 1
\nAnswer:
\na. 4, – 1<\/p>\n

Question 18.
\nIf f(x) = x2<\/sup> – 1 the value of f(2) is
\na. 1
\nb. 3
\nc. 4
\nd. 0
\nAnswer:
\nb. 3<\/p>\n

Question 19.
\nIf f(x) = 8 f(x) is called
\na. Constant polynomials
\nb. linear polynomials
\nc. quadratic polynomials
\nd. Cubic polynomials
\nAnswer:
\na. Constant polynomials<\/p>\n

\"KSEEB<\/p>\n

Question 20.
\nA cubic polynomial has at most
\na. 1 zeroes
\nb. 2 zeroes
\nc. 3 zeroes
\nd. 4 zeroes.
\nAnswer:
\nc. 3 zeroes<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nFind the value of P(x) = x2<\/sup> + 2x – 5 at x = 1
\nAnswer:
\nP(x) = x2<\/sup> + 2x – 5
\nP (1) = (1)2<\/sup> + 2(1) – 5 = 1 + 2 – 5
\n= 3 – 5
\nP(1) = – 2<\/p>\n

Question 2.
\nIf x = 1 is a zero of the Polynomial f(x) = x3<\/sup> – 2x2<\/sup> + 4x + K, write the value of K.
\nAnswer:
\nf(x) = x3<\/sup> – 2x2<\/sup> + 4x + K x = 1
\nf(1) = (1)3<\/sup> – 2(1)2<\/sup> + 4(1) + K
\nf(1) = 0 zero of the polynomial
\n0 = 1 – 2 + 4 + K
\n0 = 3 + K
\nK = – 3<\/p>\n

Question 3.
\nIf \u03b1 and \u03b2 are zeroes of Polynomials P(x) = x2<\/sup> – 5x + 6, then find the value of \u03b1 + \u03b2 – 3\u03b1\u03b2.
\nAnswer:
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}=\\frac{-(-5)}{1}\\) = 5 and
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{6}{1}\\) = 6
\n\u03b1 + \u03b2 – 3\u03b1\u03b2
\n= 5 – 3(6)
\n= 5 – 18 = -13
\n\u03b1 + \u03b2 – 3\u03b1\u03b2 = – 13<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nFind the zeroes of the Polynomial P(x) = 4x2<\/sup> – 12x + 9
\nAnswer:
\n\"KSEEB
\nP(x) = 4x2<\/sup> – 12x + 9
\n4x2<\/sup> – 6x – 6x + 9 = 0
\n2x(2x – 3) – 3(2x – 3) = 0
\n(2x – 3) (2x – 3) = 0
\nx = \\(\\frac{3}{2}\\), \\(\\frac{3}{2}\\)<\/p>\n

Question 5.
\nIf 1 is a zero of the Polynomial P(x) = ax2<\/sup> – 3(a – 1) x – 1, then find the value of a.
\nAnswer:
\nP(x) = ax2<\/sup> – 3(a – 1) x – 1
\n\u2234 x = 1
\nP(1) = a(1)2<\/sup> – 3(a – 1) 1 – 1
\nP(1) = a – 3a + 3 – 1
\n0 = – 2a + 2
\n2a = 2
\na = \\(\\frac{2}{2}\\) = 1
\na = 1<\/p>\n

Question 6.
\nWrite the degree of the Polynomial 3x3<\/sup> – x4<\/sup> + 5x + 3
\nAnswer:
\n3x3<\/sup> – x4<\/sup> + 5x + 3
\n– x4<\/sup> + 3x3<\/sup> + 5x + 3
\n\u2234 degree of the Polynomial is 4<\/p>\n

Question 7.
\nWrite the standard form of a cubic Polynomial.
\nAnswer:
\nax3<\/sup> + bx3<\/sup> + cx + d.<\/p>\n

Question 8.
\nFind the zero of the Polynomial , P(x) = a2<\/sup>x, a \u2260 0
\nAnswer:
\nP(x) = a2<\/sup>x
\na2<\/sup>x = 0
\nx = \\(\\frac{0}{a^{2}}\\)
\nx = 0<\/p>\n

Question 9.
\nIf \u03b1 and \u03b2 are the zeroes of the polynomial 4x2<\/sup> + 3x + 7, then find the value of \\(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\)
\nAnswer:
\n\"KSEEB<\/p>\n

III. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nFind the zeroes of the following quadratic polynomials x2<\/sup> + 4x + 4.
\nAnswer:
\nf(x) = x2<\/sup> + 4x + 4
\n0 = x2<\/sup> + 2x + 2x + 4
\n0 = x(x + 2) + 2(x + 2)
\n(x + 2) (or) (x + 2) = 0
\nx + 2 = 0 (or) x + 2 = 0
\nx = – 2 x = – 2.
\nZeroes of the Polynomials x2<\/sup> + 4x + 4 are – 2 and – 2<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nIf x = 1 is a zero of the polynomial f(x) = x3<\/sup> – 2x2<\/sup> + 4x + K, find the value of K.
\nAnswer:
\nf(x) = x3<\/sup> – 2x2<\/sup> + 4x + K
\n\u2234 x = 1
\nf(x) = (1)3<\/sup> – 2(1)2<\/sup> + 4(1) + K
\n0 = 1 – 2(1) + 4 + K [f(1) = 0]
\n0 = 1 – 2 + 4 + K.
\nK + 3 = 0
\nK = – 3<\/p>\n

Question 3.
\nFor what value of K, – 4 is a zero of the polynomial x2<\/sup> – x – (2K + 2)?
\nAnswer:
\nf(x) = x2<\/sup> – x – (2K + 2) and f(x) = 0
\nx = – 4
\nf(- 4) = (- 4)2<\/sup> (- 4) – (2K + 2)
\n0 = 16 + 4 – 2K – 2
\n20 – 2 – 2K = 0 \u21d2 18 = 2K
\nK = \\(\\frac{18}{2}\\) \u21d2 K = 9<\/p>\n

Question 4.
\nDivide p(x) by g(x) in each of the following cases and verify division algorithm.
\nP(x) = x2<\/sup> + 4x + 4, g(x) = x + 2
\nAnswer:
\nx2<\/sup> + 4x + 4 \u00f7 x + 2
\n\"KSEEB
\nq(x) = x + 2 and r(x) = 0<\/p>\n

Question 5.
\nOn dividing the polynomial. P(x) = x3<\/sup> – 3x2<\/sup> + x + 2 by a polynomial g(x) the quotient and remainder were (x – 2) and (- 2x + 4) respectively. Find g(x).
\nAnswer:
\nP(x) = x3<\/sup> – 3x2<\/sup> + x + 2
\nq(x) = (x – 2); r(x) = (- 2x + 4)
\ng(x) = ?
\n\"KSEEB
\ng(x) = x2<\/sup> – x + 1; r(x) = 0
\n\u2234 The divisor is x2<\/sup> – x + 1<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhat must be subtracted from x3<\/sup> + 5x2<\/sup> + 5x + 8 so that the resulting polynomial is exactly divisible by x2<\/sup> + 3x – 2?
\nAnswer:
\nP(x) = x3<\/sup> + 5x2<\/sup> + 5x + 8
\ng(x) = x2<\/sup> + 3x – 2
\nP(x) = g(x) q(x) + r(x)
\n\"KSEEB
\nq(x) = x + 2; r(x) = x + 12
\nIf we subtract x + 12 from x3<\/sup> + 5x2<\/sup> + 5x + 8 it will be exactly divisible by x2<\/sup> + 3x – 2<\/p>\n

Question 7.
\nWhat should be added to (x4<\/sup> – 1) so that it’ is exactly divisible by (x2<\/sup> + 2x + 1)?
\nAnswer:
\nP(x) = x4<\/sup> – 1 g(x) = x2<\/sup> + 2x + 1
\nx4<\/sup> + 0x3<\/sup> + 0x3<\/sup> + 0x – 1
\n\"KSEEB
\nr(x) = – 4x – 4
\n– {r(x)} = {4x + 4}<\/p>\n

Question 8.
\nIf the polynomials 2x3<\/sup> + ax2<\/sup> + 3x – 5 and x3<\/sup> + x2<\/sup> – 4x – a leave the same remainder when divided by x – 1 find the value of a.
\nAnswer:
\nx – 1 = 0 [\u2234 g(x) = (0)]
\nx = 1
\nP(x) = 2x3<\/sup> + ax2<\/sup> + 3x – 5
\nP(1) = 2(1)3<\/sup> + a(1)3<\/sup> + 3(1) – 5
\nP(1) = 2 + a + 3 – 5
\nP(1) = a \u279d (1)
\ng(x) = x3<\/sup> + x2<\/sup> – 4x – a
\ng(1) = (1)3<\/sup> + (1)2<\/sup> – 4(1) – a
\n= 1 + 1 – 4 – a
\ng(x) = – 2 – a \u279d (2)
\ng(1) = g(1) [\u2234 Polynomials are leave same remainder]
\na = – 2 – a
\na + a = – 2
\n2a = – 2
\na = \\(\\frac{-2}{2}\\)
\na = – 1<\/p>\n

Question 9.
\nIf (x3<\/sup> + ax2<\/sup> – bx + 10) is divisible by x3<\/sup> – 3x + 2, find the values of a and b.
\nAnswer:
\nP(x) = x3<\/sup> + ax2<\/sup> – bx + 10
\ng(x) = x2<\/sup> – 3x + 2
\nx2<\/sup> – 2x – x + 2 = 0
\nx(x – 2) – 1(x – 2) = 0
\n(x – 2) (x – 1) = 0
\nx = 2 and x = 1
\nP(x) = x3<\/sup> + ax2<\/sup> – bx + 10
\nP(x) = (1)3<\/sup> + a(1)2<\/sup> – b(1) + 10
\n= 1 + a – b + 10
\n0 = a – b + 11
\na – b = – 11 \u279d (1)
\nP(2) = (2)3<\/sup> + a(2)2<\/sup> – b(2) + 10
\nP(2) = 8 + 4a – 2b – 10
\n0 = 4a – 2b + 18
\n4a – 2b = – 18 \u279d (2)
\nMultiply equation (1) by 4 and subtract with equation (2)
\n\"KSEEB
\nb = 13
\nPut b = 13 in equation (1)
\na – b = – 11
\na – 13 = – 11
\na = – 11 + 13
\na = 2<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIf both x – 2 and x – \\(\\frac{1}{2}\\) are factors of ax2<\/sup> + 5x + b show that a = b.
\nAnswer:
\nP(x) = ax2<\/sup> + 5x2<\/sup> + b
\nx – 2 = 0
\nx = 2
\nP(x) = ax2<\/sup> + 5x + b
\nP(2) = a(2)2<\/sup> + 5(2) + b
\n0 = 4a + 10 + b \u279d (1)
\nP(x) = ax2<\/sup> + 5x + b
\nx – \\(\\frac{1}{2}\\) = 0
\nx = \\(\\frac{1}{2}\\)
\nP(x) = ax2<\/sup> + 5x + b
\n\"KSEEB
\na = – 2 \u279d (3)
\nPut a = – 2 in equation (1)
\n4a – 10 + b = 0
\n4(- 2) + 10 + b = 0
\n– 8 + 10 + b = 0
\n2 + b = 0
\nb = -2 \u279d (4)
\nFrom (3) and (4)
\na = b.<\/p>\n

Question 11.
\nIf \u03b1 and \u03b2 are the zeroes of the quadratic polynomials f(x) = 2x2<\/sup> – 5x + 7, find a polynomial whose zeroes are 2\u03b1 + 3\u03b2 and 3\u03b1 + 2\u03b2.
\nAnswer:
\nSince \u03b1 and \u03b2 are the zeroes of the quadratic polynomial f(x) = 2x2<\/sup> – 5x + 7.
\n\"KSEEB
\nLet S and P denote respectively the sum and product of the zeroes of the required polynomial.
\nThen, S = (2\u03b1 + 3\u03b2) + (3\u03b1 + 2\u03b2)
\n= 5(\u03b1 + \u03b2) = 5 \u00d7 \\(\\frac{5}{2}=\\frac{25}{2}\\) and
\nP = (2\u03b1 + 3\u03b2) (3\u03b1 + 2\u03b2)
\n\u21d2 P = (6\u03b12<\/sup> + 6\u03b22<\/sup> + 13\u03b1\u03b2)
\n= 6\u03b12<\/sup> + 12\u03b1\u03b2 + \u03b1\u03b2 + 6\u03b22<\/sup>
\n\u21d2 P = 6(\u03b12<\/sup> + \u03b22<\/sup> + 2\u03b1\u03b2 + \u03b1\u03b2
\n= 6(\u03b1 + \u03b2)2<\/sup> + \u03b1\u03b2
\n= 6(\u03b12<\/sup> + \u03b22<\/sup> + 2\u03b1\u03b2) + \u03b1\u03b2
\n= 6(\u03b1 + \u03b2)2<\/sup> + \u03b1\u03b2
\n\"KSEEB
\nHence the required polynomial g(x) is given by
\ng(x) = K(x2<\/sup> – 5x + P)
\nor g(x) = K\\(\\left(x^{2}-\\frac{25}{2} x+41\\right)\\), where K and any non zero real number.<\/p>\n

Question 12.
\nIf \u03b1 and \u03b2 are the zeroes of the polynomial. 6y2<\/sup> – 7y + 2, find a quadratic polynomial whose zeroes are \\(\\frac{1}{\\alpha}\\) and \\(\\frac{1}{\\beta}\\).
\nAnswer:
\nLet P(y) = 6y2<\/sup> – 7y + 2
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nIf one zero of the polynomial 3x2<\/sup> – 8x + 2k + 1 and seven times the other, find the value of k.
\nAnswer:
\nLet \u03b1 and \u03b2 be the zeroes of the polynomial. Then as per question \u03b2 = 7\u03b1.
\nNow sum of zeroes = \u03b1 + \u03b2 = \u03b1 + 7\u03b1
\n\"KSEEB<\/p>\n

Question 14.
\nIf one zero of polynomial (a2<\/sup> + 9)x2<\/sup> + 13x + 6a and reciprocal of the other, find the value of a.
\nAnswer:
\nLet one zero of the given polynomial be \u03b1
\nThen, the other zero and \\(\\frac{1}{\\alpha}\\)
\n\u2234 Product of zeroes = \u03b1 \u00d7 \\(\\frac{1}{\\alpha}\\) = 1
\nBut, as per the given polynomial product of Zeroes = \\(\\frac{6 a}{a^{2}+9}\\)
\n\u2234 \\(\\frac{6 a}{a^{2}+9}\\) = 1 \u21d2 a2<\/sup> + 9 = 6a
\n\u21d2a2<\/sup> – 6a + 9 = 0 \u21d2 (a – 3)2<\/sup> = 0
\n\u21d2 a – 3 = 0 \u21d2 a = 3
\nHence a = 3.<\/p>\n

Question 15.
\nFind the zeroes of the polynomial f(x) = x3<\/sup> – 5x2<\/sup> – 2x + 24 if it is given that the product of its two zeroes is 12.
\nAnswer:
\nLet \u03b1, \u03b2 and \u03b3 be the zeroes of polynomial f(x) such that \u03b1\u03b2 = 12
\n\"KSEEB
\nPutting \u03b1\u03b2 = 12 in \u03b1\u03b2\u03b3 = – 24 we get
\n12\u03b3 = – 24 \u21d2 \u03b3 = \\(\\frac{-24}{12}\\) = – 2
\nNow \u03b1 + \u03b2 + \u03b3 = 5 \u21d2 \u03b1 + \u03b2 – 2 = 5
\n\u21d2 \u03b1 + \u03b2 = 7
\n\u21d2 \u03b1 = 7 – \u03b2
\n\u2234 \u03b1\u03b2 =12
\n\u21d2 (7 – \u03b2) \u03b2 = 12
\n\u21d2 7\u03b2 = \u03b22<\/sup> = 12
\n\u21d2 \u03b22<\/sup> – 7\u03b2 + 12 = 0
\n\u21d2 \u03b22 – 3\u03b2 – 4\u03b2 – 12 = 0
\n\u21d2\u03b2(\u03b2 -3) – 4(\u03b2 – 3) = 0
\n\u21d2(\u03b2 -4)(\u03b2 – 3) = 0
\n\u21d2 \u03b2 = 4 or \u03b2 = 3
\n\u03b1 = 3 or \u03b1 = 4.<\/p>\n

Question 16.
\nIf \u03b1, \u03b2, \u03b3 be zeroes of polynomial 6x3<\/sup> + 3x2<\/sup> – 5x +1, then find the value of \u03b1– 1<\/sup> + \u03b2– 1<\/sup> + \u03b3– 1<\/sup>
\nAnswer:
\nP(x) = 6x3<\/sup> + 3x2<\/sup> – 5x + 1 so
\na = 6, b = 3, c = – 5, d = 1
\n\u2234 \u03b1, \u03b2 and \u03b3 are zeroes of the polynomial p(x)
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 9 Polynomials Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Additional Questions I. Multiple Choice Questions: Question 1. The degree of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25300"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25300"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25300\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25300"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25300"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25300"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}