Advertisements
Advertisements
प्रश्न
Using the Factor Theorem, show that (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
उत्तर
Let f(x) = 2x3 + 5x2 – 28x – 15
x + 5 = 0 `\implies` x = –5
∴ Remainder = f(–5)
= 2(–5)3 + 5(–5)2 – 28(–5) – 15
= –250 + 125 + 140 – 15
= –265 + 265
= 0
Hence, (x + 5) is a factor of f(x).
Now, we have,
2x2 – 5x – 3
`x + 5")"overline(2x^3 + 5x^2 - 28x - 15)`
2x3 + 10x2
– –
– 5x2 – 28x
– 5x2 – 25x
+ +
– 3x – 15
– 3x – 15
+ +
0
∴ 2x3 + 5x2 – 28x – 15 = (x + 5)(2x2 – 5x – 3)
= (x + 5)[2x2 – 6x + x – 3]
= (x + 5)[2x(x – 3) + 1(x – 3)]
= (x + 5)(2x + 1)(x – 3)
APPEARS IN
संबंधित प्रश्न
If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
If x + a is a common factor of expressions f(x) = x2 + px + q and g(x) = x2 + mx + n; show that : `a = (n - q)/(m - p)`
What should be subtracted from 3x3 – 8x2 + 4x – 3, so that the resulting expression has x + 2 as a factor?
Find the value of a , if (x - a) is a factor of x3 - a2x + x + 2.
Show that (x – 1) is a factor of x3 – 5x2 – x + 5 Hence factorise x3 – 5x2 – x + 5.
Show that (2x + 1) is a factor of 4x3 + 12x2 + 11 x + 3 .Hence factorise 4x3 + 12x2 + 11x + 3.
If (2x + 1) is a factor of 6x3 + 5x2 + ax – 2 find the value of a.
If two polynomials 2x3 + ax2 + 4x – 12 and x3 + x2 – 2x + a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.
If mx2 – nx + 8 has x – 2 as a factor, then ______.
