Odisha State Board BSE Odisha 8th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(d) Textbook Exercise Questions and Answers.
BSE Odisha Class 8 Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(d)
Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) 3 × (………) = 12x
(ii) 2x × (………) = 12x²
(iii) 4 × (……….) = – 16 x²
(iv) -3x × (……..) = 15x²
ସମାଧାନ :
(i) 4x
(ii) 6x
(iii) -4x²
(iv) -5x
Question 2.
ଭାଗଫଳ ସ୍ଥିର କର ।
(i) 8x ÷ 4
(ii) 8x ÷ (-4)
(iii) (-8x) ÷ (4)
(iv) (-8x) ÷ (-4)
ସମାଧାନ :
ଭାଜକକୁ ଯେଉଁ ସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନକଲେ ଆମେ ଭାଜ୍ୟ ପାଇବା, ତାହା ହିଁ ଭାଗଫଳ ।
(i) 2x
(ii) -2x
(iii) -2x
(iv) 2x
Question 3.
ଭାଗଫଳ ସ୍ଥିର କର ।
(i) 21x² ÷ 3
(ii) -21x² ÷ 3x
(iii) 21x² ÷ (-7x)
(iv) 21x² ÷ 3x²
(v) 21x² ÷ (-3x²)
ସମାଧାନ :
(i) 7x²
(ii) -7x
(iii) -3x
(iv) 7
(v) -7
Question 4.
ଭାଗଫଳ ସ୍ଥିର କର ।
(i) (15x² + 10) ÷ 5
(ii) (16x² – 12) ÷ 4
(iii) (24x² – 8x + 12) ÷ 4
(iv) (20x² + 15x) ÷ 5x
(v) (24x² +20) ÷ 4x
(vi) (48x² – 44x) ÷ (-4x)
ସମାଧାନ :
(i) (15x² + 10) ÷ 5 = \(\frac{15 x^2+10}{5}=\frac{15 x^2}{5}+\frac{10}{5}\) = 3x² + 2
(ii) (16x² – 12) ÷ 4 = \(\frac{16 x^2+12}{4}=\frac{16 x^2}{4}+\frac{12}{3}\) = 4x² – 3
(iii) (24x² – 8x + 12) ÷ 4 = \(\frac{24x^2 – 8x + 12}{4}=\frac{24 x^2}{4}-\frac{8x}{4}+\frac{12}{4}\) = 6x² – 2x + 3
(iv) (20x² + 15x) ÷ 5x = \(\frac{20 x^2+15x}{5x}=\frac{20 x^2}{5x}+\frac{15x}{5x}\) = 4x + 3
(v) (24x² +20) ÷ 4x = \(\frac{24x^2+20}{4x}=\frac{24 x^2}{4x}+\frac{20x}{4x}+\frac{12}{4}\) = 6x + 5
(vi) (48x² – 44x) ÷ (-4x) = \(\frac{48x^2-44x}{-4x}=\frac{48 x^2}{-4x}-\frac{44x}{-4x}+\frac{12}{4}\) = -12x + 11