10th 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’. 2x2 + 3x
+ 2 ,y2 + 2
Þ Any quadratic polynomial in x is of the form ax2 +
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 – x3, x3, √2 x3.
Þ General form of a cubic polynomial is ax3 +
bx2 + 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 ax2 +
bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola
representing y = ax2 + 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) = ax2 + b
x + c, a ≠ 0, then you know that x – α
and x – β are the factors of p(x).
Therefore, ax2 + bx + c = a(x – α) (x – β)= ax2 – a(α + β)x + a α β]
Therefore, ax2 + bx + c = a(x – α) (x – β)= ax2 – a(α + β)x + a α β]
Comparing the coefficients of x2, 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
ax3 + bx2 + c x + d= a(x-a)(x-b)(x-g) = ax3 – a(a+b+g)x2 + 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 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).
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 x2 +
7x + 10, and verify the relationship between the zeroes and the coefficients.
Q.2. Find the zeroes of the polynomial x2 – 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) = 3x3 – 5x2 –
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 2x4 – 3x3 –
3x2 + 6x – 2, if you know that two of its zeroes are √2 and − √2
Q.7. Obtain all other zeroes of 3x4 + 6x3 –
2x2 – 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 x3 –
3x2 + x + 1 are a – b, a, a + b,
find a and b.
Q.10. If two zeroes of the polynomial x4 –
6x3 – 26x2 + 138x –
35 are 2 ± √3 , find other zeroes.
Q.11. If the polynomial x4 –
6x3 + 16x2 – 25x + 10
is divided by another polynomial x2 – 2x + 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 f(x)
= x2 - p (x+1) - c, Show that (α+ 1) (β + 1) = 1- c.
Q.13. 1. For which values of a and b , are the zeros of g(x) = x3+2x2+a, also the zeros of the polynomial f(x) =x5-x4-4x3+3x2+3x+b ? Which zeros of f(x) are not the zeros of g(x)?
Q.13. 1. For which values of a and b , are the zeros of g(x) = x3+2x2+a, also the zeros of the polynomial f(x) =x5-x4-4x3+3x2+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 3x2-
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 2x2+2ax+5x+10then find the value of a. [Ans : 2]
Q.15. If two zeros of the polynomials x4 – 6x3 – 26x2
+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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