Advertisements
Advertisements
प्रश्न
Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorize the given polynomial completely.
उत्तर
f(x) = 5x3 + 4x2 – 5x – 4
Let 5x + 4 = 0, then 5x = -4
⇒ x = `(-4)/(5)`
∴ `f(-4/5) = 5(-4/5)^3 + 4(-4/5)^2 -5(-4/5) -4`
= `5 xx (-64/125) + 4 xx (16)/(25) + 4 - 4`
= `-(64)/(25) + (64)/(25) + 4 - 4` = 0
∵ `f(-4/5)` = 0
∴ (5x + 4) is a factor of f(x)
Now dividing f(x) by 5x + 4, we get
5x3 + 4x2 – 5x – 4
= (5x + 4)(x2 – 1)
= (5x + 4)[(x)2 – (1)2]
= (5x + 4)(x + 1)(x – 1)
`5x + 4")"overline(5x^3 + 4x^2 - 5x - 4)("x^2 - 1`
5x3 + 4x2
– –
–5x – 4
–5x – 4
+ +
x
APPEARS IN
संबंधित प्रश्न
Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = 0.
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Using Remainder Theorem, factorise : x3 + 10x2 – 37x + 26 completely.
If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Using the Reminder Theorem, factorise of the following completely.
2x3 + x2 – 13x + 6
If x – 2 is a factor of each of the following three polynomials. Find the value of ‘a’ in each case:
x2 - 3x + 5a
If x – 2 is a factor of each of the following three polynomials. Find the value of ‘a’ in each case:
x3 + 2ax2 + ax - 1
While factorizing a given polynomial, using remainder and factor theorem, a student finds that (2x + 1) is a factor of 2x3 + 7x2 + 2x – 3.
- Is the student's solution correct stating that (2x + 1) is a factor of the given polynomial?
- Give a valid reason for your answer.
Also, factorize the given polynomial completely.
If f(x) = 3x + 8; the value of f(x) + f(– x) is ______.
