**Master Purushottam**

Let f(a) = 2a2 + 2xa + 5x + 10

As (a + x) is a factor of 2a2 + 2xa + 5x + 10, f(-x) = 0

So, f(-x) = 2x2 – 2x2 – 5x + 10 = 0

Or, -5x + 10 = 0

Thus, x = 2

*-Answered by Master Purushottam* On 07 October 2019:09:09:32 PM

**Question 1.** **Asked on :**23 March 2020:07:33:34 AM

Find the remainder when x*x*x+3x*x+3x+1 is divided by x+1?

*-Added by Aman Senapati* **Mathematics** » **POLYNOMIALS**

**Question 2.** **Asked on :**23 March 2020:07:27:04 AM

Divide (x*x*x)+1 by x+1

*-Added by Aman Senapati* **Mathematics** » **POLYNOMIALS**

**Question 3.** **Asked on :**07 October 2019:11:36:04 AM

**Find the value of “x” in the equation 2a2 + 2xa + 5x + 10 if (a + x) is one of its factors.**

*-Added by Himanshi Verma* **Mathematics** » **POLYNOMIALS**

**Answer:**

Let f(a) = 2a2 + 2xa + 5x + 10

As (a + x) is a factor of 2a2 + 2xa + 5x + 10, f(-x) = 0

So, f(-x) = 2x2 – 2x2 – 5x + 10 = 0

Or, -5x + 10 = 0

Thus, x = 2

*-Answered by Master Purushottam* On 07 October 2019:09:09:32 PM

**Question 4.** **Asked on :**07 October 2019:11:34:52 AM

**How many zeros does the polynomial (x – 3)2 – 4 can have? Also, find its zeroes.**

*-Added by Himanshi Verma* **Mathematics** » **Polynomials**

**Answer:**

Given equation is (x – 3)2 – 4

Now, expand this equation.

=> x2 + 9 – 6x – 4

= x2 – 6x + 5

As the equation has a degree of 2, the number of zeroes it will have is 2.

Now, solve x2 – 6x + 5 = 0 to get the roots.

So, x2 – x – 5x + 5 = 0

=> x(x-1) -5(x-1) = 0

=> (x-1)(x-5)

So, the roots are 1 and 5.

*-Answered by Himanshi Verma* On 07 October 2019:11:36:47 AM

**Question 5.** **Asked on :**07 October 2019:11:32:08 AM

** Find the quadratic polynomial if its zeroes are 0, √5.**

*-Added by Himanshi Verma* **Mathematics** » **Polynomials**

**Answer:**

A quadratic polynomial can be written using the sum and product of its zeroes as:

x2 +(α + β)x + αβ = 0

Where α and β are the roots of the equation.

Here, α = 0 and β = √5

So, the equation will be:

x2 +(0 + √5)x + 0(√5) = 0

Or, x2 + √5x = 0

*-Answered by Himanshi Verma* On 07 October 2019:11:34:36 AM

**Question 6.** **Asked on :**07 October 2019:11:29:31 AM

** Find the value of “p” from the equation x2 + 3x + p, if one of the zeroes of the polynomial is 2.**

*-Added by Himanshi Verma* **Mathematics** » **Polynomials**

**Answer:**

As 2 is the zero of the polynomial,

x2 + 3x + p, for x = 2

Now, put x = 2

22 + 3(2) + p = 0

=> 4 + 6 + p = 0

Or, p = -10

*-Answered by Himanshi Verma* On 07 October 2019:11:30:47 AM

**Question 7.** **Asked on :**17 August 2019:05:57:38 PM

Find a quadratic polynomial , The sum and product of whose zeroes are -3 and 2 respectively.

*-Added by Khushi Chauhan* **Mathematics** » **POLYNOMIALS **

**Answer:**

Let the quadratic polynomial be ax^{2}+bx+c=0 and its zeroes be α and β.

sum of zeroes α+β = -3 = -b/a

product of zeroes αβ = 2= c/a

If a=1 then b=3, c=2

So, the quadratic polynomial is **x ^{2}+3x+2.**

*-Answered by Himanshi Verma* On 18 August 2019:11:00:35 AM

**Question 8.** **Asked on :**19 April 2019:08:23:46 AM

Check whether the first polynomials is a factors of the second polynomials by dividing the second polynomials by the first polynomials.

*-Added by Prince Rawat* **Mathematics** » **POLYNOMIALS**

**Answer:**

what is the polynomials.

*-Answered by Akki chauhan* On 24 August 2019:03:13:10 PM

**Question 9.** **Asked on :**27 January 2019:11:08:39 AM

(4-k)x^{2}+(2k+4)x+(8k+1)=0.

k=?

एक पुर्ण वर्ग है ?

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

The given quadratic will be a perfect square if it has two real and equal roots. To have two real and equal roots the discriminant must be zero.

i.e., b²-4ac=0

or, (2k+4)²-4(4-k)(8k+1)=0

or, 4k²+16k+16-4(32k-8k²+4-k)=0

o², 4k²+16k+16-128k+32k²-16+4k=0

or, 36k²-108k=0

or, 36k(k-3)=0

either 36k=0

or, k=0

or, k-3=0

or, k=3

∴, k=0,3 Ans.

*-Answered by Himanshi Verma* On 02 February 2019:01:43:20 PM

**Question 10.** **Asked on :**27 January 2019:10:59:13 AM

abx^{2}_{+(b}^{2}_{ac)x-bc=0.सरल करे|}

_{}

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

Consider, abx +(b-ac)x-bc = 0

⇒ abx + bx – acx – bc = 0⇒ bx (ax + b) – c(ax + b) = 0⇒ (ax + b)(bx – c) = 0*-Answered by Himanshi Verma* On 02 February 2019:01:44:13 PM

**Question 11.** **Asked on :**27 January 2019:10:54:59 AM

ax^{2}_{+4ax+(a}^{2}_{-b}^{2}_{)=0}

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

Jul 16, 2018 - 4x² +4bx -(a² -b²) = 0 4x² +4bx -a² + b² =0 (2x)² + 2.(2x).b + b² -a² =0 (2x + b)² -a² = 0 use formula, a²-b² = (a -b)(a + b) {2x + b -a}{2x + b +a} =0

*-Answered by priyanshu kumar* On 09 October 2019:11:29:56 AM

**Question 12.** **Asked on :**27 January 2019:10:51:05 AM

1/a+b+x=1/a+1/b+1/x,a+b not equal 0.सरल करे |

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

ON solving the above equation we have a quadratic equation

(a+b)x^2+(a+b)^2x+ab(a+b)

on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)

we got -a and -b as roots.

According to me this is much easier then others when compared.

*-Answered by Himanshi Verma* On 10 March 2019:04:49:29 PM

**Answer:**

what is your question

*-Answered by Akki chauhan* On 30 August 2019:08:53:43 PM

**Question 14.** **Asked on :**12 January 2019:10:49:51 AM

Factorise:

x^{3}-23x^{2}+142x-120

*-Added by Akki chauhan* **Mathematics** » **Polynomials**

**Answer:**

x^{3 }-23x^{2}+142x-120

=x^{3}-x^{2}-22x^{2}+22x+120-120

=x^{2}(x-1)-22x(x-1)+120(x-1)

=(x-1)(x^{2}-22x+120)

=(x-1)(x^{2}-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

*-Answered by Shivang Gupta* On 25 August 2019:05:08:24 PM

**Question 15.** **Asked on :**12 January 2019:10:45:37 AM

If the polynomials ax^{3}+3x^{2}-13 and 2x^{3}-5x+a , are divided by x+2 if the remainder in each case is the same, then find the value of 'a'

*-Added by Akki chauhan* **Mathematics** » **Polynomials**

**Answer:**

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

=a*8+3*4-13

=8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

=2*8-10+a

=16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

*-Answered by Shivang Gupta* On 25 August 2019:05:08:54 PM

**Question 16.** **Asked on :**12 January 2019:10:38:50 AM

If a = 9-4√5, then find the value of

a^{2} + 1a^{2}

*-Added by Akki chauhan* **Mathematics** » **Polynomials**

**Answer:**

(a+1/a)2=a2+1/a2+2

a2+1/a2=(a+1/a)2-2→ I

Now,

a=9-4root5

1/a=1/9-4root5

=(1/9-4root5)×9+4root5/9+4root5

=(9+4root5)

a+1/a=9-4root5+9+4root5

=81

Now,

a2+1/a2=(a+1/a)2-2 →(from I)

=(81)2-2

=6561-2

=6559

*-Answered by Nishant Verma* On 17 October 2019:02:43:54 PM

**Question 17.** **Asked on :**11 January 2019:10:04:29 AM

What is Degree of the Polynomials?

*-Added by Prince Rawat* **Mathematics** » **POLYNOMIALS**

**Answer:**

Highest power of X in algebraic expressions is called the degree of polynomial.

*-Answered by Nishant Verma* On 17 October 2019:02:42:40 PM

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