{"id":6354,"date":"2023-05-21T16:43:50","date_gmt":"2023-05-21T11:13:50","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=6354"},"modified":"2023-05-22T17:30:24","modified_gmt":"2023-05-22T12:00:24","slug":"ap-ssc-10th-class-maths-solutions-chapter-3-intext-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-ssc-10th-class-maths-solutions-chapter-3-intext-questions\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 3 Polynomials InText Questions"},"content":{"rendered":"

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials InText Questions<\/h2>\n

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 3 Polynomials InText Questions and Answers.<\/p>\n

10th Class Maths 3rd Lesson Polynomials InText Questions and Answers<\/h3>\n

Do these<\/span><\/p>\n

Question 1.
\nState which of the following are polynomials and which are not? Give reasons.\u00a0 \u00a0(Page No. 48)
\n(i) 2x3<\/sup>
\n(ii) \\(\\frac{1}{x-1}\\)
\n(iii) 4z2<\/sup> + \\(\\frac{1}{7}\\)
\n(iv) m2<\/sup> – \u221a2 m + 2
\n(v) p-2<\/sup> + 1
\nAnswer:
\ni) 2x3<\/sup> is a polynomial.
\nii) \\(\\frac{1}{x-1}\\) is not a polynomial because its power is negative integer exponent.
\niii) 4z2<\/sup> + \\(\\frac{1}{7}\\) is a polynomial.
\niv) m2<\/sup> – \u221a2 m + 2 is a polynomial.
\nv) p-2<\/sup> + 1 is not a polynomial because its power is negative integer exponent.<\/p>\n

\"AP<\/p>\n

Question 2.
\np(x) = x2<\/sup> – 5x – 6, find the values of p(l), p(2), p(3), p(0), p(-l), p(-2), p(-3).\u00a0 \u00a0 (Page No. 49)
\nAnswer:
\nGiven polynomial p(x) = x2<\/sup> – 5x – 6
\np(1) = (1)2<\/sup> – 5(1) – 6 = 1 – 5 – 6 = – 10
\np(2) = (2)2<\/sup> – 5(2) – 6 = 4 – 10 – 6 = -12
\np(3) = 32<\/sup> – 5(3) – 6 = 9 – 15 – 6 = – 12
\np(0) = 02<\/sup> – 5(0) – 6 = – 6
\np(-1) = (-1)2<\/sup> – 5(-1) – 6 = 1 + 5 – 6 = 0
\np(-2) = (-2)2<\/sup> – 5(-2) – 6 = 4 + 10 – 6 = 8
\np(-3) = (-3)2<\/sup> – 5(-3) – 6 = 9 + 15 – 6 = 18<\/p>\n

Question 3.
\np(m) = m2<\/sup> – 3m + 1, find the values of p(1)and p(-1).\u00a0 \u00a0 (Page No. 49)
\nAnswer:
\nGiven polynomial p(m) = m2<\/sup> – 3m + 1
\np(1) = (1)2<\/sup> – 3(1) + 1 = 1 – 3 + 1 = 2 – 3 = – 1
\np(-1) = (-1)2<\/sup> – 3(-1) + 1 = 1 + 3 + 1 = 5<\/p>\n

Question 4.
\nLet p(x) = x2<\/sup> – 4x + 3. Find the values of p(0), p(l), p(2), p(3) and obtain zeroes of the polynomial p(x).\u00a0 \u00a0(Page No. 50)
\nAnswer:
\nGiven polynomial p(x) = x2<\/sup> – 4x + 3
\np(0) = (0)2<\/sup> – 4(0) + 3 = 3
\np(1) = (1)2<\/sup> – 4(1) + 3 = 1 – 4 + 3 = 0
\np(2) = (2)2<\/sup> – 4(2) + 3 = 4 – 8 + 3 = – 1
\np(3) = (3)2<\/sup> – 4(3) + 3 = 9 – 12 + 3 = 0
\nWe see that p(1) and p(3) are zeroes of the polynomial p(x).<\/p>\n

\"AP<\/p>\n

Question 5.
\nCheck whether -3 and 3 are the zeroes of the polynomial x2<\/sup> – 9.\u00a0 \u00a0 (Page No. 50)
\nAnswer:
\nGiven polynomial p(x) = x2<\/sup> – 9
\nZero of the polynomial p(x) = 0
\nx2<\/sup> – 9 = 0
\n\u21d2 x2<\/sup> = 9
\n\u21d2 x = V9 = \u00b1 3
\n\u2234 x = + 3, – 3
\n\u2234 Zeroes of the polynomial p(x) are – 3 and 3.<\/p>\n

Try these<\/span><\/p>\n

Question 1.
\nWrite 3 different quadratic, cubic and 2 linear polynomials with different number of terms. (Page No. 48)
\nAnswer:
\nQuadratic polynomials:
\n\"AP
\nCubic polynomials :
\n\"AP
\nLinear polynomials :
\nf(t) = \u221a2 t + 5
\ng(u) = \\(\\frac{2}{3}\\) u – \\(\\frac{5}{2}\\) and
\nq(y) = 3y
\nYes, we can write polynomials of any degree.<\/p>\n

Question 2.
\nWrite a quadratic polynomial and a cubic polynomial in variable x in the general form. (Page No. 49)
\nAnswer:
\nGeneral form of a quadratic polynomial having variable \u2018x\u2019 is
\nf(x) = ax3<\/sup> + bx2<\/sup> + c, a \u2260 0
\nGeneral form of a cubic polynomial having variable ‘x’ is
\nf(x) = ax3<\/sup> + bx2<\/sup> + cx + d, a \u2260 0<\/p>\n

\"AP<\/p>\n

Question 3.
\nWrite a general polynomial q(z) of degree n with coefficients that are b0<\/sub>…. bn<\/sub>. What are the conditions on b0<\/sub>…. bn<\/sub>. (Page No. 49)
\nAnswer:
\nq(z) = b0<\/sub>zn<\/sup> + b1<\/sub>zn-1<\/sup> + b2<\/sub>zn-2<\/sup> …….. + bn-1<\/sub>z + bn<\/sub> is a polynomial of n degree where b0<\/sub>, b1<\/sub>, b2<\/sub>,…… bn-1<\/sub>, bn<\/sub> are real coefficients and b0<\/sub> \u2260 0.<\/p>\n

Do this<\/span><\/p>\n

Question 1.
\nDraw the graph of i) y = 2x + 5, ii) y = 2x – 5, iii) y = 2x and find the point of intersection on X – axis. Is the x-coordinates of these points also the zero of the polynomial?\u00a0 \u00a0(Page No. 52)
\nAnswer:
\ni) Given that y = 2x + 5
\n\"AP
\n\"AP
\nResult: The graph y = 2x + 5 cuts the X – axis at the point (-2.5, 0).
\nHence, the zeroes of the polynomial is -2.5.<\/p>\n

ii) Given that y = 2x – 5
\n\"AP
\n\"AP
\nResult: The graph of y = 2x – 5 cuts the X – axis at the point (2.5, 0).
\nThe zeroes of the polynomial is 2.5 = \\(\\frac{5}{2}\\)<\/p>\n

\"AP<\/p>\n

iii) Given that y = 2x
\n\"AP
\n\"AP
\nResult: The graph passes through the origin.
\nSo, the zeroes of the polynomial y = 2x is zero.<\/p>\n

Try these<\/span><\/p>\n

Question 1.
\nDraw the graphs of (i) y = x2<\/sup> – x – 6 (ii) y = 6 – x – x2<\/sup> and find zeroes in each case. What do you notice? (Page No. 53)
\nAnswer:
\ni) Given that y = x2<\/sup> – x – 6
\n\"AP
\n\"AP
\nResult: From the graph we observe that 3 and -2 are the intersecting points of X – axis.
\nSo, the zeroes of given quadratic polynomial are 3 and -2.<\/p>\n

\"AP<\/p>\n

ii) Given that y = 6 – x – x2<\/sup>
\n\"AP
\n\"AP
\nResult: From the graph we observe that – 3 and 2 are the intersecting points ol X – axis.
\nSo, the zeroes of given quadratic polynomial are – 3 and 2.<\/p>\n

Question 2.
\nWrite three quadratic polynomials that have 2 zeroes each.\u00a0 \u00a0(Page No. 55)
\nAnswer:
\ny = x2<\/sup> – x – 2 having two zeroes, i.e., (2, 0) and (- 1, 0).
\ny = 3 – 2x – x2<\/sup> having two zeroes i.e., (1,0) and (- 3, 0).
\ny = x2<\/sup> – 3x – 4 having two zeroes i.e., (-1, 0) and (4, 0)<\/p>\n

\"AP<\/p>\n

Question 3.
\nWrite one quadratic polynomial that has one zero. (Page No. 55)
\nAnswer:
\nQuadratic Polynomial y = x2<\/sup> – 6x + 9 has only one zero i.e., 3.<\/p>\n

Question 4.
\nHow will you verify if a quadratic polynomial it has only one zero?\u00a0 (Page No. 55)
\nAnswer:
\nIf the graph of the given quadratic polynomial touches X – axis at exactly one point, then I can confirm it has only one zero.<\/p>\n

Question 5.
\nWrite three quadratic polynomials that have no zeroes for x that are real numbers.\u00a0 (Page No. 55)
\nAnswer:
\nThe quadratic polynomials y = 2x2<\/sup> – 4x + 5 and y = – 3x2<\/sup> + 2x – 1 and y = x2<\/sup> – 2x + 4 have no zeroes.<\/p>\n

Question 6.
\nFind the zeroes of cubic polynomials
\n(i) – x3<\/sup>
\n(ii) x2<\/sup> – x3<\/sup>
\n(iii) x3<\/sup> – 5x2<\/sup> + 6x
\nwithout drawing the graph of the polynomial.\u00a0 (Page No. 57)
\nAnswer:
\ni) Given polynomial is y = – x3<\/sup>
\nf(x) = -x3<\/sup> ; f(x) = 0
\nx3<\/sup> = 0
\nx = \\(\\sqrt[3]{0}\\) = 0
\n\u2234 Zero of the polynomial f(x) is only one i.e., 0.<\/p>\n

\"AP<\/p>\n

ii) Given that y = x2<\/sup> – x3<\/sup>
\nf(x) = x2<\/sup> (1 – x)
\nf(x) = 0
\n\u21d2 x2<\/sup> (1 – x) = 0
\n\u21d2 x2<\/sup> = 0 and 1 – x = 0
\n\u21d2 x = 0 and x = 1
\n\u2234 The zeroes of the polynomial f(x) are two i.e., 0 and 1.<\/p>\n

iii) Given that x3<\/sup> – 5x2<\/sup> + 6x Let f(x) = x3<\/sup> – 5x2<\/sup> + 6x
\n= x(x2<\/sup> – 5x + 6)
\n= x(x2<\/sup> – 2x – 3x + 6)
\n= x[x(x – 2) – 3(x – 2)]
\n= x(x – 2) (x – 3)
\n\u2234 The zeroes of the polynomial f(x) are x = 0 and x = 2 and x = 3<\/p>\n

Do these<\/span><\/p>\n

Question 1.
\nFind the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial. (Page No. 62)
\ni) p(x) = x2<\/sup> – x – 6
\nii) p(x) = x2<\/sup> – 4x + 3
\niii) p(x) = x2<\/sup> – 4
\niv) p(x) = x2<\/sup> + 2x + 1
\nAnswer:
\ni) Given polynomial p(x) = x2<\/sup> – x – 6
\nWe have x2<\/sup> – x – 6 = x2<\/sup> – 3x + 2x – 6
\n= x(x – 3) + 2(x – 3)
\n= (x – 3) (x + 2)
\nSo, the value of x2<\/sup> – x – 6 is zero when x – 3 = 0 or x + 2 = 0
\ni.e., x = 3 or x = -2
\nSo, the zeroes of x2<\/sup> – x – 6 are 3 and – 2.
\n\u2234 Sum of the zeroes = 3 – 2 = 1
\n= – \\(\\frac{\\text { Coefficient of } x}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{-(-1)}{1}\\) = 1
\nAnd product of the zeroes = 3 \u00d7 (-2) = -6
\n= \\(\\frac{\\text { Constant term }}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{-6}{1}\\) = -6<\/p>\n

ii) p(x) = x2<\/sup> – 4x + 3
\nAnswer:
\nGiven polynomial p(x) = x2<\/sup> – 4x + 3
\nWe have, x2<\/sup> – 4x + 3 = x2<\/sup> – 3x – x + 3
\n= x(x – 3) – 1 (x – 3)
\n= (x – 3) (x – 1)
\nSo, the value of x2<\/sup> – 4x + 3 is zero when x – 3 = 0 or x – 1 =0, i.e.,
\nwhen x = 3 or x = 1 So, the zeroes of x2<\/sup> – 4x + 3 are 3 and 1
\n\u2234 Sum of the zeroes = 3 + 1 = 4
\n= – \\(\\frac{\\text { Coefficient of } x}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{-(-4)}{1}\\) = 4
\nAnd product of the zeroes = 3 \u00d7 1 = 3
\n= \\(\\frac{\\text { Constant term }}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{3}{1}\\) = 3<\/p>\n

\"AP<\/p>\n

iii) Given polynomial p(x) = x2<\/sup> – 4
\nWe have, x2<\/sup> – 4 = (x – 2) (x + 2)
\nSo, the value of x2<\/sup> – 4 is zero
\nwhen x – 2 = 0 or x + 2 = 0
\ni.e., x = 2 or x = – 2
\nSo the zeroes of x2<\/sup> – 4 are 2 and – 2
\n\u2234 Sum of the zeroes = 2 + (- 2) = 0
\n= – \\(\\frac{\\text { Coefficient of } x}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{-0}{1}\\) = 0
\nAnd product of the zeroes = 2 \u00d7 (-2) = -4
\n= \\(\\frac{\\text { Constant term }}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{-4}{1}\\) = -4<\/p>\n

iv) Given polynomial p(x) = x2<\/sup> + 2x + 1
\nWe have x2<\/sup> + 2x + 1 = x2<\/sup> + x + x + 1
\n= x(x + 1) + l(x + 1)
\n= (x + 1) (x + 1)
\nSo, the value of x2<\/sup> + 2x + 1 is zero
\nwhen x + 1 = 0 (or) x + 1 = 0, i.e.,
\nwhen x = – 1 or – 1
\no, the zeroes of x2<\/sup> + 2x + 1 are – 1 and – 1.
\n\u2234 Sum of the zeroes = (-1) + (-1) = -2
\n= – \\(\\frac{\\text { Coefficient of } x}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{-2}{1}\\) = -2
\nAnd product of the zeroes = (-1) \u00d7 (-1) = 1
\n= \\(\\frac{\\text { Constant term }}{\\text { Coefficient of } x^{2}}\\) = \\(\\frac{1}{1}\\) = 1<\/p>\n

Question 2.
\nIf \u03b1, \u03b2 and \u03b3 are the zeroes of the given cubic polynomials, find the values as given in the table. (Page No. 66)
\n\"AP
\nAnswer:
\nl) Given polynomial is x3<\/sup> + 3x2<\/sup> – x – 2.
\nComparing given polynomial with ax3<\/sup> + bx2<\/sup> + cx + d,
\nthe values are a = 1, b = 3, c = -l, d = -2
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{-b}{a}\\) = \\(\\frac{-3}{1}\\) = -3
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{c}{a}\\) = \\(\\frac{-1}{1}\\) = -1
\n\u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) = \\(\\frac{-(-2)}{1}\\) = 2<\/p>\n

\"AP<\/p>\n

2) Given polynomial is 4x3<\/sup> + 8x2<\/sup> – 6x – 2
\nCompare the polynomial with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nThen a = 4, b = 8, c = – 6 and d = – 2
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{-b}{a}\\) = \\(\\frac{-8}{4}\\) = -2
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{c}{a}\\) = \\(\\frac{-6}{4}\\) = \\(\\frac{-3}{2}\\)
\n\u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) = \\(\\frac{-(-2)}{4}\\) = \\(\\frac{1}{2}\\)<\/p>\n

3) Given polynomial is x3<\/sup> + 4x2<\/sup> – 5x – 2
\nCompare the polynomial with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nThen a = 1, b = 4, c = – 5 and d = – 2
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{-b}{a}\\) = \\(\\frac{-4}{1}\\) = -4
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{c}{a}\\) = \\(\\frac{-5}{1}\\) = -5
\n\u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) = \\(\\frac{-(-2)}{1}\\) = 2<\/p>\n

4) Given polynomial is x3<\/sup> + 5x2<\/sup> + 4
\nCompare the polynomial with ax3<\/sup> + bx2<\/sup> + cx + d = 0
\nThen a = 1, b = 5, c = 0 and d = 4
\n\u03b1 + \u03b2 + \u03b3 = \\(\\frac{-b}{a}\\) = \\(\\frac{-5}{1}\\) = -5
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac{c}{a}\\) = \\(\\frac{0}{1}\\) = 0
\n\u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) = \\(\\frac{-4}{1}\\) = -4<\/p>\n

\"AP<\/p>\n

Try this<\/span><\/p>\n

Question 1.
\ni) Find a quadratic polynomial with zeroes -2 and \\(\\frac{1}{3}\\). (Page No. 64)
\nAnswer:
\nLet the quadratic polynomial be ax2<\/sup> + bx + c, a \u2260 0 and its zeroes be \u03b1 and \u03b2.
\nHere \u03b1 = – 2 and \u03b2 = \\(\\frac{1}{3}\\)
\nSum of the zeroes = \u03b1 + \u03b2
\n= -2 + \\(\\frac{1}{3}\\) = \\(\\frac{-5}{3}\\)
\nProduct of the zeroes = \u03b1\u03b2
\n= \\(\\frac{1}{3}\\) \u00d7 (-2) = \\(\\frac{-2}{3}\\)
\n\u2234 ax2<\/sup> + bx + c is [x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2]
\n= [x2<\/sup> – \\(\\left(\\frac{-5}{3}\\right)\\)x + \\(\\left(\\frac{-2}{3}\\right)\\)]
\nthe quadratic polynomial will be 3x2<\/sup> + 5x – 2.<\/p>\n

\"AP<\/p>\n

ii) What is the quadratic polynomial whose sum of zeroes is \\(\\frac{-3}{2}\\) and the product of zeroes is -1.
\nAnswer:
\nLet the quadratic polynomial be ax2<\/sup> + bx + c and its zeroes be \u03b1 and \u03b2.
\nHere \u03b1 + \u03b2 = \\(\\frac{-3}{2}\\) and \u03b1\u03b2 = -1
\nThus, the polynomial formed = x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2
\n= x2<\/sup> – \\(\\left(\\frac{-3}{2}\\right)\\)x + (-1)
\n= x2<\/sup> + \\(\\frac{3x}{2}\\) – 1
\nThe other polynomials are (x2<\/sup> + \\(\\frac{3x}{2}\\) – 1)
\nthen the polynomial is 2x2<\/sup> + 3x – 2.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials InText Questions AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials InText Questions and Answers. 10th Class Maths 3rd Lesson Polynomials InText Questions and Answers Do these Question 1. State which of the following are polynomials and which are not? … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6354"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=6354"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6354\/revisions"}],"predecessor-version":[{"id":40642,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6354\/revisions\/40642"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}