Find values to fill these boxes and complete the identity: ◻ x 2 − ◻ x − ◻ = ◻ x 2 + ◻ x − 4 x − 14 = 5 x ( 2 x + 7 ) − ◻ ( 2 x + 7 ) = ( ◻ x − ◻ ) ( ◻ x + ◻ )

Given, □ x 2 - □ x - □ = □ x 2 + □ x - 4 x - 14 = 5 x ( 2 x + 7 ) - □ ( 2 x + 7 ) = ( □ x - □ ) ( □ x + □ ) Let us name the empty boxes with variables, a , b , c , d , e , f , g , h , i , and j as follows: a x 2 - b x - c = d x 2 + e x - 4 x - 14 . . . . i = 5 x 2 x + 7 - f 2 x + 7 . . . . i i = g x - h i x + j . . . . i i i Considering L.HS. a x 2 - b x - c = d x 2 + e x - 4 x - 14 = d x 2 + e x - 2 2 x + 7 . . . . . i v = 5 x ( 2 x + 7 ) - 2 ( 2 x + 7 ) (given) . . . . v = 5 x - 2 2 x + 7 . . . . . v i Now, comparing equations i i , and v we have f = 2 , g = 5 , h = 2 , i = 2 , j = 7 From R.H.S. of equation v we have 5 x 2 x + 7 - 2 2 x + 7 = 10 x 2 + 35 x - 4 x - 14 . . . . v i i From equations i v , and v i i we have d = 10 , and e = 35 Since a x 2 - b x - c = d x 2 + e x - 4 x - 14 = 10 x 2 + 35 x - 4 x - 14 = 10 x 2 + 31 x - 14 ⇒ a = 10 , b = - 31 , and c = 14 Hence, the values in the empty boxes are as follows: 10 x 2 − - 31 x − 14 = 10 x 2 + 35 x − 4 x − 14 = 5 x ( 2 x + 7 ) − 2 ( 2 x + 7 ) = ( 5 x − 2 ) ( 2 x + 7 )

<span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">If one zeros of the polynomial ( a 2 + 9 ) x 2 + 13 x + 6 a <span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> is the reciprocal of the other, find the value of a .

Let α and 1 α be the zeros of ( a 2 + 9 ) x 2 + 13 x + 6 a . Then, we have α × 1 α = 6 a a 2 + 9 ⇒ 1 = 6 a a 2 + 9 ⇒ a 2 + 9 = 6 a ⇒ a 2 − 6 a + 9 = 0 ⇒ ( a − 3 ) 2 = 0 ⇒ a − 3 = 0 ⇒ a = 3

E M B I B E

Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 79

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IMPORTANT

Earn 100

A polynomial is divided by $(x - 1) .$ The quotient obtained is $3 x^{3} - x^{2} - x - 4$ , and the remainder is $- 5 .$ Which polynomial meets these conditions?

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Important Questions on Polynomials

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10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 80

Simplification of

(x + y)^{3} - (x - y)^{3}

is

k x^{2} y + 2 y^{3}

. Find the value of

k

.

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Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 81

If two zeroes of a cubic polynomial are zero, then it does not contain linear and constant terms.

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Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 82

Find the value of $x,$ if $a + 1 = 0$

and $x^{2} + ax - 6 = 0 .$ .

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10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 83

Solve the given equation using factorisation method.

$4 (2 x - 3)^{2} - (2 x - 3) - 14 = 0$

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10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 84

Roopa picked $24$ rose flowers from her garden and distributed them equally among her friends. Had there been $2$ more friends, each would have received $1$ flower less. Find the number of friends of Roopa.

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Mathematics Textbook of Competency Based Questions for Class X>Chapter 2 - Polynomials>Exercise>Q 85

The difference between the sides of a right-angled triangle containing the right angle is