### X maths Chapter: 02 Polynomials CBSE Test paper Trend Setter problems

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^{th}Polynomials Ch-02-Mathematics [Key points]
Þ Any algebraic expression having non zero integral
power (whole number) is called polynomial.

Þ If p(x) is a polynomial in x, the highest power of x in p(x) is
called the degree of the polynomial p(x).

Þ A polynomial of degree 1 is called a linear polynomial. For
example, 2x – 3, √3x+1,y +√3

Þ A polynomial of degree 2 is called a quadratic polynomial.
The name ‘quadratic’ has been derived from the word ‘quadrate’, which means
‘square’. 2x

^{2}+ 3x + 2 ,y^{2}+ 2
Þ Any quadratic polynomial in x is of the form ax

^{2}+ bx + c, where a, b, c are real numbers and a ≠ 0.
Þ A polynomial of degree 3 is called a cubic polynomial e.g.
2 –

*x*^{3},*x*^{3}, √2*x*^{3}.
Þ General form of a cubic polynomial is ax

^{3}+ bx^{2}+ c x + d, where, a, b, c, d are real numbers and a ≠ 0.
Þ If p(x) is a polynomial in x, and if k is any real number, then
the value obtained by replacing x by k in p(x), is called the value of p(x) at
x = k, and is denoted by p(k). A real number k is said to be a zero of a
polynomial p(x), if p (k) = 0.

Þ Thus, the zero of a linear polynomial is related to its
coefficients because if

*k*is a zero of*p*(*x*) =*ax*+*b*, then*p*(*k*) = a k + b = 0, i.e., k = -b/a
Þ Zero of a linear polynomial ax + b is −b/a

ÞThe graph of y = ax + b is a straight line like the graph of y =
ax + b is a straight line passing through the points (– 2, –1) and (2, 7) and
straight line straight line intersects the x-axis at exactly one point

ÞThe linear polynomial ax + b, a ≠ 0, has exactly one zero,
namely, the x-coordinate of the point (-b/a , 0) where the graph of y = ax + b
intersects the x-axis.

Þ The graph of equation y = ax+ + b
x + c has one of the two shapes either open upwards like È (a >
0 ) or open downwards like Ç ( a < 0) . These curves are
called

**parabolas**.
Þ The zeroes of a quadratic polynomial ax

^{2}+ bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola representing y = ax^{2}+ bx + c intersects the x-axis.
Þ In general, given a polynomial p(x) of degree n, the graph
of y = p(x) intersects the x- axis at atmost n[n or
less than n] points.

Þ A polynomial p(x) of degree n has at most n zeroes.

Þ if α and β are the zeroes of the quadratic
polynomial p(x) = ax

Therefore, ax

^{2}+ b x + c, a ≠ 0, then you know that x – α and x – β are the factors of p(x).Therefore, ax

^{2}+ bx + c = a(x – α) (x – β)= ax^{2}– a(α + β)x + a α β]
Comparing the coefficients of x

^{2}, x and constant terms on both the sides, we get , a = k**,**b = – k(α + β) and c = kαβ.
This
gives

**α + β = -b/a ; α β = c/a**

Þ Relationship between the zeroes of a cubic polynomial and
its coefficients of

ax

^{3}+ bx^{2}+ c x + d= a(x-a)(x-b)(x-g) = ax^{3}– a(a+b+g)x^{2 }+ a(ab +bg+ga) - aabg
Comparing the coefficients of terms on both the sides, we
get, α + β + γ = –b/a; α
β + β γ + γ α =c/a; α β γ =– d/a

Þ Division algorithm states that given any polynomial

Þ Division algorithm states that given any polynomial

*p*(*x*) and any non-zero polynomial*g*(*x*), there are polynomials*q*(*x*) and*r*(*x*) such that*p*(*x*) =*g*(*x*)*q*(*x*) +*r*(*x*), where*r*(*x*) = 0 or degree*r*(*x*) < degree*g*(*x*).
Check point [Formative Assessment]

Q. 1. Find the zeroes of the quadratic polynomial x^{2}+ 7x + 10, and verify the relationship between the zeroes and the coefficients.

Q.2. Find the zeroes of the polynomial x

^{2}– 3 and verify the relationship between the zeroes and the coefficients.
Q.3. Find a quadratic polynomial, the sum and product of whose
zeroes are – 3 and 2, respectively.

Q.4. Verify that 3, –1, -1/3 are the zeroes of the
cubic polynomial p(x) = 3x

^{3 }– 5x^{2}– 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Q.5. Find a quadratic polynomial if the sum and product of its
zeroes respectively √2, 1/3

Q.6. Find all the zeroes of 2x

^{4 }– 3x^{3}– 3x^{2}+ 6x – 2, if you know that two of its zeroes are √2 and − √2
Q.7. Obtain all other zeroes of 3x

^{4}+ 6x^{3}– 2x^{2}– 10x – 5, if two of its zeroes are √(5/3) , -√(5/3)
Q.8. Find a cubic polynomial with the sum, sum of the product of
its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14
respectively

Q.9. If the zeroes of the polynomial

*x*^{3}– 3*x*^{2}+*x*+ 1 are*a*–*b*,*a*,*a*+*b*, find*a*and*b*.
Q.10.

**If two zeroes of the polynomial***x*^{4}– 6*x*^{3}– 26*x*^{2}+ 138*x*– 35 are 2 ± √3 , find other zeroes.
Q.11.

**If the polynomial***x*^{4}– 6*x*^{3}+ 16*x*^{2}– 25*x*+ 10 is divided by another polynomial*x*^{2}– 2*x*+*k*, the remainder comes out to be*x*+*a*, find*k*and*a*.
Q.12. If α and β are the zeros of the quadratic polynomial

Q.13. 1. For which values of a and b , are the zeros of g(x) = x

*f(x) = x*, Show that (α+ 1) (β + 1) = 1- c.^{2}- p (x+1) - cQ.13. 1. For which values of a and b , are the zeros of g(x) = x

^{3}+2x^{2}+a, also the zeros of the polynomial f(x) =x^{5}-x^{4}-4x^{3}+3x^{2}+3x+b ? Which zeros of f(x) are not the zeros of g(x)?(Ans. 1 and 2 are the zeros of g(x) which are not the zeros f(x) and this happens when a= -2 , b= -2)

Q.14. 1. If one zero of the polynomial 3x

^{2}- 8x –( 2k + 1) is seven times the other , find both zeroes of the polynomial and the value of k .
( Ans. The zeroes of the given polynomial are 1/3 , 7/3 and the value of k is (-5/3)

Q.14. if (x+a) is a
factor of 2x

^{2}+2ax+5x+10then find the value of a. [Ans : 2]
Q.15. If two zeros of the polynomials x

^{4}– 6x^{3}– 26x^{2}+138x -35 are 2- Root3 and 2+ root3 , find all the zeros .
(
Ans. Zeros p(x) are 2 - root3 and 2 + root3 , -5 and 7)

Q.16. It is true to
say that for k= 2 the pair of linear equation 3x+y = 1;(2k-1)x + (k-1)y = 2k+1
has no solution. Justify

Ans:

**Given**system of equations is:
3x+y=1 and

(2k-1)x+(k-1)y=2k+1

They can be rewrite as:

3x + y - 1 = 0 and

(2k-1)x + (k-1)y - (2k+1) = 0

We know that, for the given system to have no solution, a1/a2 =
b1/b2 ¹ c1/c2

Þ 3/(2k-1) = 1/(k-1)⇒

⇒ 3k - 3 = 2k - 1

⇒ k = 3-1

⇒ k = 2

Hence, for k = 2, the given system of equation will have no solution.

Ans: Let the number of questions correctly answered be x

Therefore, number of questions incorrectly answered will be 120 - x.

According to the given condition,

1× x – (1/2)(120-x) = 90

⇒ x - 60 +(1/2)( x = 90

⇒ (3/2) x = 150

⇒ x= 100

Class X Polynomial Test Paper-1

Class X Polynomial Test Paper-2

Class X Polynomial Test Paper-3

Class X Polynomial Test Paper-4

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