Polynomials
Mathematics Class Tenth
NCERT Exercise 2.4 (Optional)
Question: 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between zeroes and the coefficients in each case.
(i) `2x^3+x^2-5x+2`; `1/2, 1, -2`
Solution:
Given, `p(x)=2x^3+x^2-5x+2`
And zeroes of this polynomial are `1/2, 1, -2`
Verification of zeroes of the given polynomial
Now,
`p(1/2) = 2(1/2)^3+(1/2)^2-5(1/2)+2`
`=2xx1/8+3xx1/4-5/2+2`
`=1/4+1/4-5/2+2`
`=(1+1-10+8)/4`
`=0/4 = 0`
And, `p(1) = 2xx1^3+1^2-5xx1+2`
`=2+1-5+2`
` =5-5 =0`
And, `p(-2) = 2(-2)^3+(-2)^2-5(-2)+2`
`=-16+4+10+2 =0`
Thus, `1/2, 1` and `-2` are the zeroes of the given polynomial.
Verification of the relationship between zeroes and coefficient of the given polynomial
Let, a general polynomial `ax^3+bx^2+cx+d`
By comparing the given polynomial with `ax^3+bx^2+cx+d`, we get
`a = 2, b=1, c=-5` and `d=2`
Thus, we can take `alpha = 1/2, beta = 1`, and `gamma = -2`
Sum of zeroes
`=alpha + beta + gamma = 1/2+1+(-2)`
`=(-1)/2 = (-b)/a`
`alpha beta + beta gamma + alpha gamma = 1/2xx1+1(-2)+1/2(-2)`
`=(-5)/2 = c/a`
And product of zeroes,
`alpha beta gamma = 1/2xx1xx(-2) = (-1)/1`
`=(-2)/2 = (-d)/a`
Therefore, relationship between the zeroes and the coefficients is verified.
(ii) `x^3-4x^2+5x-2`; `2, 1, 1`
Solution:
Given, `p(x) = x^3-4x^2+5x-2`
And zeroes of the polynomial are `2, 1, 1`
Verification of zeroes,
Now,
`p(2) = 2^3-4(2)^2+5(2)-2`
`= 8 -4xx4+10-2`
`=8-16+10 -2`
`=0`
And, `p(1) = 1^3-4(1)^2+5(1)-2`
`= 1-4+5-2 = 6-6`
`=0`
Thus, 2, 1, 1 are the zeroes of the given polynomial.
Verification of the relationship between zeroes and coefficient of the given polynomial.
By comparing the given polynomial with `ax^3+bx^2+cx+d`, we get
`a = 1, b = -4, c = 5` and `d = -2`
Sum of zeroes `=2+1+1 = 4`
`=(-(-4))/1 = (-b)/a`
Addition zeroes by multiplying two zeroes at a time
`(2)(1)+(1)(1)+(2)(1) = 2+1+2`
`=5 = 5/1 = c/a`
Multiplication of zeroes,
`2xx1xx1 = 2`
`=(-(-2))/1 = (-d)/a`
Hence, the relationship between zeroes and the coefficients is verified.
Question: 2. Find a cubic polynomial with the sum, sum of the product of its zeroes are taken two at a time, and the product of its zeroes as `2, -7, -14` respectively.
Solution:
Let a cubic polynomial is `ax^3+bx+cx+d`, and their zeroes are `alpha,\ beta` and `gamaa`
Given, `alpha + beta + gamma = 2 = 2/1 = (-b)/a`
And, `alpha\ beta + beta\ gamma + alpha\ gamma = -7 = (-7)/1=c/a`
And, `alpha\ beta\ gamma = -14 = -14/1 = (-d)/a`
Therefore, `a=1, b = -2, c = -7` and `d = 14`
Thus, the polynomial so formed is after substituting the values of `a,\ b,\ c` and `d` is `x^3-2x^2-7x+14` Answer
Question: 3. If the zeroes of the polynomial `x^3-3x^2+x+1` and are `a-b,\ a, \ a+b`, find `a` and `b`.
Solution:
Given, `p(x) = x^3-3x^2+x+1`
And given, zeroes are `a-b, a` and `a+b`
Thus, After comparing given polynomial with `px^3+qx^2+rx+t`, we get
`p=1`, `q=-3`, `r=1` and `t=1`
Now, sum of zeroes,
`=a-b+a+a+b = 3a = (-q)/p`
After putting values of `q` and `p`, we get
`(-(-3))/1 = 3a`
`=>3 = 3a`
`:. a = 3/1=1`
Now, after replacing the values of `a` we get that zeroes are
`1-b, 1, 1+b`
Now, Multiplication of zeroes
`= 1(1-b)(1+b) = (-t)/p`
`=>(1-b)(1+b) = (-t)/p`
`=>1-b^2 = (-t)/p`
After substituting values of `t` and `p`, we get
`1-b^2 = (-1)/1`
`=> 1-b^2 = -1`
`=>-b^2 =-1-1`
`=>b^2 = 1+1 =2`
`=>b= (+-sqrtr)`
Therefore, `a=1` and `b = sqrt2` or `-sqrt2` Answer
Question: 4. If two zeroes of the polynomial `x^4-6x^3-26x^2+138x-35` are `2+-sqrt3`. Find other zeroes.
Solution:
Given, two zeroes of the polynomials are `2+sqrt3` and `2-sqrt3`
Therefore, `(2+sqrt3)(2-sqrt3)` is a factor of given polynomial.
Thus, `[x-(2+sqrt3)][x-(2-sqrt3)]`
`=x^2-2x-\cancel(sqrt3\x)-2x+\cancel(sqrt3\x)+2^2-(sqrt3)^2`
`=x^2-4x+4-3`
`=x^2 -4x+1`.
Thus, `x^2-4x+1` is a factor of the given polynomial.
Now, dividing the given polynomial with this factor `x^2-4x+1` will give the other factor of polynomial.
Thus,
`x^4-6x^3-26x^2+138x-35`
`=(x^2-4x+1)(x^2-2x-35)`
This, means `(x^2-2x-35)` is also a factor of given polynomial.
Now, `(x^2-2x-35) `
`=x^2 +5x-7x-35`
`=x(x+5)-7(x+5)`
`=(x-7)(x+5)`
Thus, the value of the polynomial is also zero when `x-7 = 0` or `x+5=0`
Therefore,
Now, if `x-7 = 0`
Therefore, `x= 7`
And if `x+5=0`
Therefore, `x = -5`
Thus, 7 and `-5` are the other zeroes of the given polynomial. Answer
Question: 5. If the polynomial `x^4-6x^3+16x^2-25x+10` is divided by another polynomial `x^2-2x+k`, the remainder comes out to be `x+a`, find `k` and `a`.
Solution:
Given, Dividend `=x^4-6x^3+16x^2-25x+10`
Divisor `=x^2-2x+k`
Remainder `=x+a`
We know that,
Dividend = Divisor `xx` Quotient + Remainder
∴ Dividend – Remainder = Divisor × Quotient
`=> x^4-6x^3+16x^2-25x+10-x-a` `=x^4-6x^3+16x^2-26x+10-a`
Thus, `x^4 - 6x^3+16x^2 - 26x+10-a` will be exactly divisible by `x^2-2x+k`
Since, `x^4 - 6x^3+16x^2 - 26x+10-a` is exactly divisible by `x^2-2x+k`, thus,
Remainder `(-10+2k)x+(10-a-8k+k^2)` will be equal to zero.
i.e. `(-10+2k)x = 0`
And `(10-a-8k+k^2)` will also equal to zero (0)
Thus, if `(-10+2k)x = 0`
`:. -10+2k = 0`
`=>2k = 10`
`=> k = 10/2 = 5`
And, if `10-a-8k+k^2 = 0`
After substituting the value of `k=5`, we get
`10-a-8xx5+5^2 =0`
`=> 10-a-40+25 =0`
`=> 35-40-a=0`
`=>-5 - a =0`
`=>-a = 5`
`=>a = -5`
Thus, `k=5` and `a = -5` Answer
Summary: Polynomials
(1) Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials respectively.
(2) A quadratic polynomial in `x` with real coefficients is of the form `ax^2+bx+c` where `a,b, c` are real numbers and `a!=0`.
(3) The zeroes of a polynomial `p(x)` are precisely the `x-`coordinates of the points, where the graph of `y=p(x)` intersects the `x-`axis.
(4) A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes.
(5) If `alpha` and `beta` are the zeroes of the quadratic polynomial `ax^3+bx+c`, then
`alpha + beta = -(b)/a` and `alpha\ beta = c/a`
(6) If `alpha,\ beta,\ gamma` are the zeroes of the cubic polynomial `ax^3+bx^2+cx+d`, then
`alpha+beta+gamma = (-b)/a`
And, `alpha\ beta + beta\ gamma + gamma\ alpha = c/a`
And, `alpha\ beta\ gamma = (-d)/a`
(7) The 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) `
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