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The expression that should be subtracted from the polynomial f(x) = x^{4} + 2x^{3}13x^{2} – 12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x^{2} – 4x + 3 is
Stepbystep explanation:
Given: Dividend =
Divisor = x^{2}  4x + 3
Solution :
Since we know that :
Since the remainder is 2x3
So, 2x3 must be subtracted from
so that the resulting polynomial is exactly divisible by x^{2}  4x + 3
If (x + 1) is a factor of x^{2} – 3ax + 3a – 7, then the value of a is
Since (X+1) is a factor of x2  3ax + 3a  7
Therefore , x+1=0
or, x=1
Putting the value of x in the the p(x) = x2  3ax + 3a  7
or, (1)^{2} 3(a)(1)+3a7=0
1+3a+3a7=0
6a=6
or, a=1
Therefore the value of a=1.
The value of quadratic polynomial f (x) = 2x^{2}– 3x 2 at x = 2 is ……
We put the value x=2 in the polynomial 2x^{2}3x2
f(2)=2(2)^{2}3(2)2=2*4+62=12
If the degree of the dividend is 5 and the degree of the divisor is 3, then the degree of the quotient will be
The degree of the divisor is 2
Stepbystep explanation:
x^{5}÷x^{3}=x^{2}
By the Division algorithm 155 = 9(17) + 2. Where remainder is 155 mod 9.
On dividing f(x) = x^{3} – 3x^{2} + x + 2 by a polynomial g(x) the quotient and remainder q(x) and r(x) are (x – 2) and (2x + 4) respectively, then g(x) is
Divisor*Quotient+Remainder= Divident
g(x)(x2)+(2x+4)=x33x2+x+2
g(x)(x2)=x33x2+x+2+2x4
g(x)(x2)=x33x2+3x2
g(x)=x33x2+3x2/(x2)
g(x)=x^{2}x+1
The three zeroes of the polynomial 2x^{3} + 5x^{2} – 28x – 15 _____
Since the solution is not a complex number, it is a real number. Since rational numbers include all integers, natural numbers and fractions , So the numbers are rational numbers.
If two of the zeroes of the polynomial f (x) = x^{4} – 3x^{3} – x^{2} + 9x – 6 are √3 and √3 then all the zeroes are
If (x + 1) is a factor of 2x^{3} + ax^{2} + 2bx + 1, then find the values of a and b given that 2a – 3b = 4.
Given is the value of a and b.
given that (x+1) is a factor of p(x)
therefore, 1 is a zero of given p(x)
p(x) = 2x3 +ax2 +2bx +1
substituting the value of 1 in the given p(x), we get
p(x)=2*(1)3 +a *(1)2 +2*b*(1)+ 1
= 2 + a 2b + 1
= 1 + a  2b
or,a  2b = 1
also given that 2a  3b = 4
so we got two equations;
a  2b = 7 ...(1)
2a  3b = 4 ...(2)
(1)* (2)* (4) 2 = 2a  4b = (3)
(2)* 1 = 2a  3b = 4 (4)
(4)  (3) = [2a  2a ] + [3b  (4b)] = 4  2
3b + 4b = 2
therefore b = 2
substituting the value of bin (3)
2a  4b = 2
2a  (4*2) = 2
2a  8 = 2
2a = 2 + 8
2 = 10
a = 10 / 2
therefore a = 5
so we get the value of a and b
that is ; a = 5 and b = 2
Which of the given is the set of zeroes of the polynomial p(x) = 2x^{3} + x^{2} – 5x + 2
We have the polynomial 2x^{3}+x25x+2
By Hit and trial method, x=1 is the solution.
Dividing the polynomial by x1
We get quotient as = 2x^{2}+3x2
The value of p when x^{3} + 9x^{2} + px – 10 is exactly divisible by (x+ 2) is ____
When the polynomial f(x) = 4x^{3} + 8x^{2} + 8x + 7 is divided by the polynomial g(x) = 2x^{2} – x + 1, the quotient and the remainder are
When x^{2} – 2x + k divides the polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10, the remainder is (x + a). The value of a is _________
Given that the remainder is (x + a)
⇒ (4k – 25 + 16 – 2k)x + [10 – k(8 – k) ] = x + a
⇒ (2k – 9)x + [10 – 8k + k^{2} ] = x + a
On comparing both the sides, we get
2k – 9 = 1
⇒ 2k = 10
∴ k = 5
Also 10 – 8k + k^{2} = a
⇒ 10 – 8(5) + 5^{2} = a
⇒ 10 – 40 + 25 = a
∴ a = – 5
If the degree of the divisor g(x) is one then the degree of the remainder r(x) is
Degree of remainder is one less than the degree of the divisor because if the remainder is of the same degree then it can be divided further , its just like when the divisor is a number the remainder is less than the number because if it is more than or equal to that number it can further be divided. So, the degree is zero.
If the polynomial (2x + 3) is a factor of the polynomial 2x^{3} + 9x^{2} – x – b. The value of b is_______
let the given polynomial be
since 2x+3=0
if (2x+3) is a factor of f(x). therefore by the factor theorem f(3/2) =0
If two of the zeros of the polynomial x^{4} – 17x^{3} + 90x^{2} – 166x + 92 are (7 + √3) and (7 – √3), then the other two zeros are _________
If f(x) is divided by g(x), g(x) ≠ 0, then there exist two polynomials q(x) and r(x) such that
Dividend = Divisor × Quotient + Remainder.
When we divide f(x) by g(x) then q(x) is quotient and r(x) is remainder.
Therefore, f(x) = g(x) × q(x) + r(x).
The expression that should be added to the polynomial f(x) = x^{4} + 2x^{3} – 2x^{2} + x + 1, so that it should be exactly divisible by (x^{2} + 2x – 3) is
When x^{3} – 3x^{2} + 5x – 3 is divided by x^{2} – k , the remainder is 7x + a . Then the value of k is_____
If two zeroes of a polynomial 4x^{4} 20x^{3} +23 x^{2} + 5x – 6 are ½ and – ½, the how many more zeroes does it have?
The polynomial with degree 4 is a biquadratic equation. And a polynomial has no. of zeros equal to the degree of the polynomial. So total no. of zeros are 4. 2 zeros are already provided to us , so it has 2 more zeros.
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