plz help me plz plz plz plz plz plz plz plz plz plz
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Sroory Dont Now
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2.p(x) = x3 - 3x2 - 9x - 5
Factors of 5 = 1,5,-1,-5
p(1) = 13 - 3 x (1)2 - 9 x (1) - 5
= 1 -3 - 9 - 5
= 1 - 17 = - 16 ≠ 0
p(-1) = (-1)3 x 3 x (-1)2 - 9 x (-1) -5
= -1 - 3 + 9 - 5 = -9 + 9 = 0
p(-1) =0
∴ (x+1) is a factor of p(x)
p(x) ÷ (x+1)
x3 - 3x2 - 9x - 5 ÷ x + 1
= x2 - 4x - 5
x2 - 4x - 5
x2 - 5x + 1x - 5
= x ( x - 5) + 1 ( x - 5)
= ( x - 5) ( x + 1)
3.Let p(x) = x³+13x²+32x+20
By trial, we find that
p( - 1) = ( - 1 {)}^{3} + 13( - 1 {)}^{2} + 32( - 1) + 20 \\ \\ = - 1 + 13 - 32 + 20 = 0
By factor theorem, x-(-1) i, e(x+1) is a factor of p(x)
Now,
{x}^{3} + 13x + 32x + 20 \\ \\ = {x}^{2}(x + 1) + 12x(x + 1) + 20(x + 1) \\ \\ = (x + 1)( {x}^{2} + 12x + 20) \\ \\ = (x + 1)( {x}^{2} + 2x + 10x + 20) \\ \\ = (x + 1)(x(x + 2) + 10(x + 2) \\ \\ = (x + 1)(x + 2)(x + 10)
4.2y³ + y² - 2y - 1
= 2y³ + 2y² -y² - y - y -1
= 2y²(y + 1) -y(y + 1) - 1(y + 1)
= (y +1 )(2y² - y - 1)
= (y + 1){2y² - 2y + y - 1}
= (y + 1){2y(y -1 ) + 1(y - 1)}
=(y +1 )(2y +1 )(y -1)
5.ANSWER
x
3
−23x
2
+142x−120
=x
3
−x
2
−22x
2
+22x+120x−120
=x
2
(x−1)−22x(x−1)+120(x−1)
=(x−1)(x
2
−22x+120)
=(x−1)[x
2
−10x−12x+120]
=(x−1)[x(x−10)−12(x−10)]
=(x−1)(x−10)(x−12)
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