प्रश्नावली-3E
1. (i) यदि द्विघात बहुपद f(x) के लिए f(4)=0 हो तो f(x) का एक गुणनखण्ड क्या होगा?
Answer:—-
x-4
(ii)यदि बहुपद f(x) के लिए f( -1 ) हो तो f(x) का
3
एक गुणनखण्ड लिखें-
Answer:—-
3x+1=0 x=-1/3
2. यदि द्विघात बहुपद p(x) के लिए p(3)=6 हो तो p(x) -6 का एक गुणनखण्ड लिखें-
Answer:—-
x-3
3.गुणनखण्ड प्रमेय की सहायता से गुणनखण्ड निकाले-
(i) y2-5y+6
माना कि p(x)=y2-5y+6
यहाँ अचर पद 6 है| अत: 6 के गुणनखण्ड
+1, +2, +3, +6
– – – –
अब y2-5y+6 में बारी बारी से y की जगह इन मानों को रखने पर हम पाते हैं कि
p(1)=(1)2-5×1+6=2
p(-1)=(-1)2-5×(-1)+6=1+5+6=12
p(2)=(2)2-5×2+6=4-10+6=0
p(-2)=(-2)2-5×(-2)+6=4+10+6=20
p(3)=(3)2-5×3+6=9-15+6=0
अत: (y-2)(y-3) गुणनखण्ड संभव है|
x3 -6×2+11x-6
putting x=1
(1)3-6×(1)2+11×1-6
1-6+11-6=0
x-1)x3-6×2+11x-6(x2-5x+6
x3 -x2
5×2+11x
-5×2+5x
6x-6
6x-6
0
x2-5x+6
x2-3x-2x+6
x(x-3)-2(x-3)
(x-3)(x-2)
3.(ii) x2+7x-18
उत्तर:—
यहाँ अचर पर -18 है| जिसके गुणनखण्ड
+1, +2, +3, +6, +9, +18
– – – – – –
p(x)=x2+7x-18
p(1)=(1)2+7×1-18
=1+7-18=8-18=-10
p(2)=(2)2+7×2-18=4+14-18
=4+14-18=0
p(3)=(3)+7×3-18=9+21-18
=30-18=12
p(6)=(6)2+7×6-18=36+42-18
78-18=60
p(9)=(9)2+7×9-18=81+63-18
144-18=126
p(-9)=(-9)2+7×(-9)-18
81-64-18=81-81=0
गुणनखण्ड (x-2)(x+9)
(iii) 6×2+17x+5
यहाँ अचर पर 5. गुणनखण्ड
+1, +5,
– –
p(x)=6×2+17x+5
p(1)=6×(1)2+17×1+5
6+17+5=28
p(-1)=6×(-1)2+17×(-1)+5=6-17+5
6×2+17x+5
6×2+15x+2x+5
3x(2x+5)+1(2x+5)
(3x+1)(2x+5)
4.(i) x3-6×2+11x-6
p(1)=(1)3-6×(1)2+11×1-6
1-6+11-6=12-12=0
x-1
x-1) x3-6×2+11x-6(x2-5x+6
x3-x2
-5×2+11x
-5×2+5x
6x-6
6x-6
x
x2-5x+6=x2-3x-2x+6
x(x-3)-2(x-3)=(x-2)(x-3)
(ii) x3+4×2+x-6
p(1)=(1)3+4×(1)2+1-6=1+4+1-6=0
x-1
x-1)x3+4×2+x-6(x2+5x+6
x3- x2
5×2+x
5×2-5x
6x-6
6x-6
X
x2+5x+6=x2+3x+2x+6
x(x+3)+2(x+3)=(x+2) (x+3)
(iii) y3-8y2+17y-10
p(1)=(1)3-8×(1)2+17×1-10
1-8+17-10=18-18=0
y-1
y-1)y3-8y2+17y-10(y2-7y+10
y3-y2
-7y2+17y
-7y2+7y
10y-10
10y-10
X
y2-7y+10=y2-5y-2y+10
y(y-5)-2(y-5)=(y-5)(y-2)
(iv) q3-7q2+14q-8
p(1)=(1)3-7×(1)2+14×1-8
1-7+14-8=15-15=0
q-1) q3-7q2+14q-8(q2-6q+8
q3 -q2
-6q2+14q
-6q2+6q
8q-8
8q-8
X
q2-6q+8=q2-4q-2q+8
q(q-4)-2(q-4)=(q-4)(q-2)
(v) x3-7x+6
p(1)=(1)3-7×1+6=7-7=0
x-1
x-1) x3-7x+6(x2+x-6
x3-x2
x2-7x
x2-x
-6x+6
-6x+6
X
x2+x-6=x2+3x-2x-6
x(x+3)-2(x+3)=(x+3)(x-2)
(vi) y3+3y2-4
p(1)=(1)3+3×(1)2-4=1+3-4=0
y-1
y-1) y3+3y2-4(y2+4y+4
y3-y2
4y2-4
4y2-4y
4y-4
4y-4
X
y2+4y+4=y2+2y+2y+4
y(y+2)+2(y+2)=(y+2)(y+2)
5.(i) z3+8z2+17z+10
p(-1)=(-1)3+8×(-1)2+17×(-1)+10
-1+8-17+10=18-18=0
z+1
z+1) z3+8z2+17z+10(z2+7z+10
z3+z2
7z2+17z
7z2+7z
10z+10
10z+10
X
z2+7z+10=z2+5z+2z+10
z(z+5)+2(z+5)=(z+5)(z+2)
(ii) x3+7×2+14x+8
p(-1)=(-1)3+7×(-1)2+14×(-1)+8
-1+7-14+8=15-15=0
x+1
x+1)x3+7×2+14x+8(x2+6x+8
x3 + x2
6×2+14x
6×2 + 6x
8x+8
8x+8
X
x2+6x+8=x2+4x+2x+8
x(x+4)+2(x+4)=(x+4) (x+2)
(iii) z3-3z2+4
p(-1)=(-1)3-3×(-1)2+4=-1-3+4=0
z+1
z+1) z3-3z2+4(z2-4z+4
z3+z2
-4z2+4
-4z2+4
X
z2-4z+4=z2-2z-2z+4
z(z-2)-2(z-2)=(z-2)(z-2)
6.(i) x3+4×2-4x-16
p(-2)=(-2)3+4×(-2)2-4×(-2)-16
-8+16+8-16=0
x+2
x+2) x3+4×2-4x-16(x2+2x-8
x3+2×2
2×2-4x
-2×2+4x
-8x-16
-8x-16
X
x2+2x-8
x2+4x-2x-8
x(x+4)-2(x+4)=(x+4)(x-2)
(ii) m3-9m2-m+105
p(-3)=(-3)3-9×(-3)2-(-3)+105
-27-81+3+105=108-108=0
m+3
m+3) m3-9m2-m+105(m2-12m+35
m3+3m
-12m2-m
-12m2-36m
35m+105
35m+105
m2-12m+35=m2-7m-5m+35
m(m-7)-5(m-7)=(m-7)(m-5)
(iii)p3+4p2-11p-30
p(-2)=(-2)3+4×(-2)2-11(-2)-30
-8+16+22-30=0
p+2
p+2) p3+4p2-11p-30(p2+2p-15
p3+2p2
2p2-11p
2p2+4p
-15p-30
-15p-30
X
p2+2p-15=p2+5p-3p-15
p(p+5)-3(p+5)=(p+5)(p-3)
7.(i) x3-2×2-x+2
p(1)=(1)3-2(1)2-1+2
1-2-1+2=0
x-1
x-1) x3-2×2-x+2(x2-x-2
x3-x2
-x2-x
-x2+x
-2x+2
-2x+2
X
x2-x-2=x2-2x+x-2
x(x-2)+1(x-2)=(x-2)(x+1)
(ii) x3+13×2+32x+20
p(-1)3=(-1)3+13×(-1)2+32×(-1)+20
-1+13-32+20=0
x+1
x+1) x3+13×2+32x+20(x2+12x+20
x3 +x2
12×2+32x
12×2+12x
20x+20
20x+20
X
x2+12x+20=x2+10x+2x+20
x(x+10)+2(x+10)=(x+10)(x+2)
(iii) x3-23×2+142x-120
p(1)=(1)3-23(1)2+142×1-120
1-23+142-120=0
x-1
x-1)x3-23×2+142x-120(x2-22x+120
x3-x2
-22×2+142x
-22×2+22x
120x-120
120x-120
X
x2-22x+120=x2-12x-10x+120
x(x-12)-10(x-12)=(x-12)(x-10)
(iv) x3+6×2+11x+6
p(-1)=(-1)3+6×(-1)2+11×(-1)+6
-1+6-11+6=0
x+1
x+1)x3+6×2+11x+6(x2+5x+6
x3+x2
5×2+11x
5×2+5x
6x+6
6x+6
X
x2+5x+6=x2+3x+2x+6
x(x+3)+2(x+3)=(x+3)(x+2)
8.(i) 12×3+4×2-3x-1
p( 1 )=
2
12×( 1 )3+4×( 1 )2-3× 1 -1
2 2 2
12× 1 +4× 1 – 3 -1
8 4 2
3 +1 – 3 -1=0
2 2
2x-1
2x-1) 12×3+4×2-3x-1(6×2+5x+1
12×3-6×2
10×2-3x
10×2-5x
2x-1
2x-1
X
6×2+5x+1=6×2+3x+2x+1
3x(2x+1)+1(2x+1)=(2x+1)(3x+1)
(ii) 8×3+12×2-2x-3
p( 1 )
2
8×( 1 )3+12×( 1 )2-2× 1 -3
2 2 2
8× 1 +12× 1 -2× 1 -3
8 4 2
1+3-1-3=0
2x-1
2x-1)8×3+12×2-2x-3(4×2+8x+3
8×3-4×2
16×2-2x
16×2-8x
6x-3
6x-3
X
4×2+8x+3=4×2+6x+2x+3
2x(2x+3)+1(2x+3)
(2x+3)(2x+1)
(iii) 6z3+13z2+2z-5
p(-1)=6×(-1)3+13×(-1)2+2×(-1)-5
-6+13-2-5=0
z+1
z+1)6z3+13z2+2z-5(6z2+7z-5
6z3+6z2
7z2+2z
7z2+7z
-5z-5
-5z-5
X
6z2+7x-5=6z2+10z-3z-5
2z(3z+5)-1(3z+5)=(3z+5)(2z-1)
(iv) 2m3-13m2+22m-8
p(2)3=2×(2)3-13×(2)2+22×2-8
2×8-13×4+44-8
16-52+44-8=0
m-2
m-2)2m3-13m2+22m-8(2m2-9m+4
2m3-4m2
-9m2+22m
-9m2+18m
4m-8
4m-8
X
2m2-9m+4=2m2-8m-m+4
2m(m-4)-1(m-4)=(m-4)(2m-1)
9. यदि बहुपद x3-ax2-13x+b के गुणनखण्ड (x-1)(x+3) हो तो a और b के मान निकाले—–
उत्तर:—
x-1=0 x+3=0
x=1 x=-3
p(x)=x3-ax2-13x+b
p(1)=(1)3-a×(1)2-13×1+b
1-a-13+b=0
-a+b=12 ——(1)
p(x)=x3-ax2-13x+b
p(-3)=(-3)3-a×(-3)2-13×(-3)+b
-27-9a+39+b=0
-9a+b=-12 ——–(2)
From eq (1)×9 & (2)
-9a+9b=108
-9a+b=-12
8b=120
b=120/8=15
putting the value of b=15 in eq (1)
-a+b=12 -a+15=12
a=15-12=3
10. यदि बहुपद 2×3+px2+qx-14 के गुणनखण्ड x-1 और x+2 हो तो p और q के मान ज्ञात करें-
उत्तर:-
x-1=0 x+2=0
x=1 x=-2
p(x)=2×3+px2+qx-14
p(1)=2×(1)3+p(1)2+q×(1)-14
2+p+q-14=0
p+q=12 ——(1)
p(x)=2×3+px2+qx-14
p(-2)=2×(-2)3+p×(-2)2+q×(-2)-14
2×(-8)+p×4-2q-14=0
4p-2q=30 ——(2)
From eq (1)×2 & 2
2p+2q=24
4p-2q=30
6p=54
p=54/6=9
Put the value of p in eq (1)
p+q=12
9+q=12
q=12-9=3
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