Advertisements
Advertisements
प्रश्न
y3 − 2y2 − 29y − 42
Advertisements
उत्तर
Let f(y) = y3 − 2y2 − 29y − 42 be the given polynomial.
Now, putting y = -2we get
`f(-2) = (-2)^ -2 (-2)^2 - 29 (-2) -42`
` = -8 -8 + 58 - 42 = -58 + 58`
` = 0`
Therefore, (y +2) is a factor of polynomial f(y).
Now,
`f(y) = y^2 (y+2) + 4y(y+2) -21(y+2)`
` = (y + 2){y^2 -4y - 21}`
` =y +2`{y^2 -7y + 3y - 21}
`=(y + 2)(y+3)(y - 7)`
Hence (y+2),(y+3) and (y - 7) are the factors of polynomial f(y).
APPEARS IN
संबंधित प्रश्न
If `x = 2` is a root of the polynomial `f(x) = 2x2 – 3x + 7a` find the value of a.
f(x) = 9x3 − 3x2 + x − 5, g(x) = \[x - \frac{2}{3}\]
Find the remainder when x3 + 3x2 + 3x + 1 is divided by \[x - \frac{1}{2}\].
Find the value k if x − 3 is a factor of k2x3 − kx2 + 3kx − k.
In the following two polynomials, find the value of a, if x − a is factor (x5 − a2x3 + 2x + a + 1).
2x4 − 7x3 − 13x2 + 63x − 45
If x + a is a factor of x4 − a2x2 + 3x − 6a, then a =
If both x − 2 and \[x - \frac{1}{2}\] are factors of px2 + 5x + r, then
Factorise the following:
z² + 4z – 12
If (x + 5) and (x – 3) are the factors of ax2 + bx + c, then values of a, b and c are
