$\frac { 2 }{ 5 } { x }^{ 5 }-\frac { 6 }{ 5 } { x }^{ 4 }-\frac { 14 }{ 15 } { x }^{ 2 }=$
$2xy-4x-3y+6=$
$25{ x }^{ 2 }-1=$
$36{ x }^{ 6 }-49=$
${ 5x }^{ 2 }-10x+5=$
${ x }^{ 2 }-6x+9=$
${ x }^{ 2 }-20x+100=$
${ x }^{ 2 }+10x+25=$
${ x }^{ 2 }+14x+49=$
${ x }^{ 3 }-4{ x }^{ 2 }+4x=$
$3{ x }^{ 7 }-27x=$
${ x }^{ 2 }-11x+30=$
$3{ x }^{ 2 }-10x+3$
$2{ x }^{ 2 }-x-1$

Solution of exercise 1

${ x }^{ 3 }+{ x }^{ 2 }$

${ x }^{ 3 }+{ x }^{ 2 }={ x }^{ 2 }(x+1)$

Solution of exercise 2

$2{ x }^{ 4 }+4{ x }^{ 2 }$

$2{ x }^{ 4 }+4{ x }^{ 2 }=2{ x }^{ 2 }({ x }^{ 2 }+2)$

Solution of exercise 3

${ x }^{ 2 }-4$

${ x }^{ 2 }-{ 2 }^{ 2 }=(x-2).(x+2)$

Solution of exercise 4

${ x }^{ 4 }-16$

${ ({ x }^{ 2 }) }^{ 2 }-{ 4 }^{ 2 }=({ x }^{ 2 }-4).({ x }^{ 2 }+4)$

Solution of exercise 5

$9+6x+{ x }^{ 2 }$

${ x }^{ 2 }+6x+9$

$a{ x }^{ 2 }+bx+c$

$a=1,\quad b=6,\quad c=9$

In order to break the middle term, you need to multiply "a" with "c":

$1\times 9=9$

Now break 9 into least possible factors:

$3\times 3=9$

The next step is to check the sign of the c and in this case, it is positive this means that you need to add the factors to come up with 6:

$3+3=6$

Now your middle term is broken, you can carry forward:

${ x }^{ 2 }+x(3+3)+9$

${ x }^{ 2 }+3x+3x+9$

$x(x+3)+3(x+3)$

$(x+3)(x+3)$

${(x+3) }^{ 2 }$

Solution of exercise 6

${ x }^{ 2 }-x-6=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=-1,\quad c=-6$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-1)+\sqrt { { (-1 })^{ 2 }-4(1)(-6) } }{ 2(1) } ,\frac { -(-1)-\sqrt { { { (-1 }) }^{ 2 }-4(1)(-6) } }{ 2(1) }$

$x=\quad \frac { 1+\sqrt { { 1 }+24 } }{ 2 } ,\frac { 1-\sqrt { { 1 }+24 } }{ 2 }$

$x=\quad \frac { 1+\sqrt { 25 } }{ 2 } ,\frac { 1-\sqrt { 25 } }{ 2 }$

$x=\quad \frac { 1+5 }{ 2 } ,\frac { 1-5 }{ 2 }$

$x=\quad \frac { 6 }{ 2 } ,\frac { -4 }{ 2 }$

$x=3,\quad x=-2$

${ x }^{ 2 }-x-6=(x-3).(x+2)$

Solution of exercise 7

${ x }^{ 4 }-10{ x }^{ 2 }+9=0$

Assuming ${ x }^{ 2 }=t$ :

${ t }^{ 2 }-10t+9=0$

$a{ t }^{ 2 }+bt+c=0$

$a=1,\quad b=-10,\quad c=9$

$t=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$t=\quad \frac { -(-10)+\sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) } ,\frac { -(-10)-\sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) }$

$t=\quad \frac { 10+\sqrt { { 100 }-36 } }{ 2 } ,\frac { 10-\sqrt { { 100 }-36 } }{ 2 }$

$t=\quad \frac { 10+\sqrt { 64 } }{ 2 } ,\frac { 1-\sqrt { 64 } }{ 2 }$

$t=\quad \frac { 10+8 }{ 2 } ,\frac { 10-8 }{ 2 }$

$t=\quad \frac { 18 }{ 2 } ,\frac { 2 }{ 2 }$

$t=9,\quad t=1$

Putting the value of t:

${ x }^{ 2 }=9,\quad { x }^{ 2 }=1$

$x=\pm 3,\quad x=\pm 1$

${ x }^{ 4 }-10{ x }^{ 2 }+9=(x+1).(x-1).(x+3).(x-3)$

Solution of exercise 8

${ x }^{ 4 }-2{ x }^{ 2 }-3=0$

Assuming ${ x }^{ 2 }=t$ :

${ t }^{ 2 }-2t-3=0$

$a{ t }^{ 2 }+bt+c=0$

$a=1,\quad b=-2,\quad c=-3$

$t=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$t=\quad \frac { -(-2)+\sqrt { { (-2) }^{ 2 }-4(1)(-3) } }{ 2(1) } ,\frac { -(-2)-\sqrt { { (-2) }^{ 2 }-4(1)(-3) } }{ 2(1) }$

$t=\quad \frac { 2+\sqrt { 4+12 } }{ 2 } ,\frac { 2-\sqrt { 4+12 } }{ 2 }$

$t=\quad \frac { 2+\sqrt { 16 } }{ 2 } ,\frac { 2-\sqrt { 16 } }{ 2 }$

$t=\quad \frac { 2+4 }{ 2 } ,\frac { 2-4 }{ 2 }$

$t=\quad \frac { 6 }{ 2 } ,\frac { -2 }{ 2 }$

$t=3,\quad t=-1$

Putting the value of t:

${ x }^{ 2 }=3,\quad { x }^{ 2 }=-1$

$x=\pm \sqrt { 3 } ,\quad x=\pm \sqrt { -1 } \notin R$

Solution of exercise 9

$2{ x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6$

$P(1)=2{ (1) }^{ 4 }+{ (1) }^{ 3 }-8{ (1) }^{ 2 }-1+6=0$

$(x-1).(2{ x }^{ 3 }+3{ x }^{ 2 }-5x-6)$

$P(1)=2{ (1) }^{ 3 }+{ 3(1) }^{ 2 }-5{ (1) }-6\neq 0$

$P(-1)=2{ (-1) }^{ 3 }+{ 3(-1) }^{ 2 }-5{ (-1) }-6=-2+3+5-6=0$

$(x-1)(x+1)(2{ x }^{ 2 }+x-6)$

$P(2)=2{ (2) }^{ 2 }+(2)-6\neq 0$

$P(-2)=2{ (-2) }^{ 2 }+(-2)-6=0$

$(x-1).(x+1).(x+2).(2x-3)$

$2x-3=2(x-\frac { 3 }{ 2 } )$

$2{ x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6=2(x-1).(x+1).(x+2).(x-\frac { 3 }{ 2 } )$

$x=1,\quad x=-1,\quad x=-2,\quad x=\frac { 3 }{ 2 }$

Solution of exercise 10

$2{ x }^{ 3 }-7{ x }^{ 2 }+8x-3$

$P(1)=2{ (1) }^{ 3 }-7{ (1) }^{ 2 }+8(1)-3=2-7+8-3=0$

$(x-1).(2{ x }^{ 2 }-5x+3)$

$P(1)=2{ (1) }^{ 2 }-5(1)+3=2-5+3=0$

$(x-1).(x-1).(2x-3)={ (x-1) }^{ 2 }.2(x-\frac { 3 }{ 2 } )=2{ (x-1) }^{ 2 }(x-\frac { 3 }{ 2 } )$

$x=\frac { 3 }{ 2 } ,\quad x=1$

Solution of exercise 11

${ x }^{ 3 }-{ x }^{ 2 }-4$

$\{ \quad \pm 1,\quad \pm 2,\quad \pm 4\quad \}$

$P(1)={ (1) }^{ 3 }-{ (1) }^{ 2 }-4=1-1-4\neq 0$

$P(-1)={ (-1) }^{ 3 }-{ (-1) }^{ 2 }-4=-1-1-4\neq 0$

$P(2)={ (2) }^{ 3 }-{ (2) }^{ 2 }-4=8-4-4=0$

$(x-2).({ x }^{ 2 }+x+2)$

${ x }^{ 2 }+x+2=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=1,\quad c=2$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(1)+\sqrt { { (1 })^{ 2 }-4(1)(2) } }{ 2(1) } ,\frac { -(1)-\sqrt { { { (1 }) }^{ 2 }-4(1)(2) } }{ 2(1) }$

$x=\quad \frac { 1+\sqrt { { 1 }-8 } }{ 2 } ,\frac { 1-\sqrt { { 1 }-8 } }{ 2 }$

$x=\quad \frac { 1+\sqrt { -7 } }{ 2 } \notin R,\frac { 1-\sqrt { -7 } }{ 2 } \notin R$

$(x-2).({ x }^{ 2 }+x+2)$

$x = 2$

Solution of exercise 12

${ x }^{ 3 }+3{ x }^{ 2 }-4x-12$

$\{ \pm 1,\quad \pm 2,\quad \pm 3,\quad \pm 4,\quad \pm 6\quad ,\pm 12\}$

$P(1)={ (1) }^{ 3 }+3{ (1) }^{ 2 }-4(1)-12=1+3-4-12\neq 0$

$P(-1)={ (-1) }^{ 3 }+{ 3(-1) }^{ 2 }-4(-1)-12=-1+3+4-12\neq 0$

$P(2)={ (2) }^{ 3 }+3{ (2) }^{ 2 }-4(2)-12=8+12-8-12=0$

$(x-2).({ x }^{ 2 }+5x+6)$

${ x }^{ 2 }+5x+6=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=5,\quad c=6$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(5)+\sqrt { { (5 })^{ 2 }-4(1)(6) } }{ 2(1) } ,\frac { -(5)-\sqrt { { { (5 }) }^{ 2 }-4(1)(6) } }{ 2(1) }$

$x=\quad \frac { -5+\sqrt { { 25 }-24 } }{ 2 } ,\frac { -5-\sqrt { { 25 }-24 } }{ 2 }$

$x=\quad \frac { -5+\sqrt { 1 } }{ 2 } ,\frac { -5-\sqrt { 1 } }{ 2 }$

$x=\quad \frac { -5+1 }{ 2 } ,\frac { -5-1 }{ 2 }$

$x=\quad \frac { -4 }{ 2 } ,\frac { -6 }{ 2 }$

$x=-2,\quad x=-3$

$(x-2).(x+2).(x+3)$

$x=2, \quad x=-2, \quad x=-3$

Solution of exercise 13

${ 6x }^{ 3 }+7{ x }^{ 2 }-9x+2$

$\{ \pm 1,\quad \pm 2,\}$

$P(1)={ 6(1) }^{ 3 }+7{ (1) }^{ 2 }-9(1)+2=6+7-9+2\neq 0$

$P(-1)={ 6(-1) }^{ 3 }+7{ (-1) }^{ 2 }-9(-1)+2=-6+7+9+2\neq 0$

$P(2)={ 6(2) }^{ 3 }+7{ (2) }^{ 2 }-9(2)+2=48+28-18+2\neq 0$

$P(-2)={ 6(-2) }^{ 3 }+7{ (-2) }^{ 2 }-9(-2)+2=-48+28-18+2=0$

$(x+2).({ 6x }^{ 2 }-5x+1)$

${ 6x }^{ 2 }-5x+1=0$

$a{ x }^{ 2 }+bx+c=0$

$a=6,\quad b=-5,\quad c=1$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-5)+\sqrt { { (5 })^{ 2 }-4(6)(1) } }{ 2(6) } ,\frac { -(-5)-\sqrt { { { (5 }) }^{ 2 }-4(6)(1) } }{ 2(6) }$

$x=\quad \frac { 5+\sqrt { { 25 }-24 } }{ 12 } ,\frac { 5-\sqrt { { 25 }-24 } }{ 12 }$

$x=\quad \frac { 5+\sqrt { 1 } }{ 12 } ,\frac { 5-\sqrt { 1 } }{ 12 }$

$x=\quad \frac { 5+1 }{ 12 } ,\frac { 5-1 }{ 12 }$

$x=\quad \frac { 6 }{ 12 } ,\frac { 4 }{ 12 }$

$x=\frac { 1 }{ 2 } \quad x=\frac { 1 }{ 3 }$

$6(x+2).(x-\frac { 1 }{ 2 } ).(x-\frac { 1 }{ 3 } )$

$x=-2,\quad x=\frac { 1 }{ 2 } ,\quad x=\frac { 1 }{ 3 }$

Solution of exercise 14

$9{ x }^{ 4 }-4{ x }^{ 2 }$

${ x }^{ 2 }(9{ x }^{ 2 }-4)$

${ x }^{ 2 }(3x+2)(3x-2)$

${ x }^{ 2 }[{ (3x) }^{ 2 }-{ (2) }^{ 2 }]$

${ x }^{ 2 }(9{ x }^{ 2 }-4)$

$9{ x }^{ 4 }-4{ x }^{ 2 }$

$x^{ 5 }+20{ x }^{ 3 }+100x$

$x(x^{ 4 }+20x^{ 2 }+100)$

$x{ (x^{ 2 }+10) }^{ 2 }$

$3x^{ 5 }-18{ x }^{ 3 }+27x$

$3x(x^{ 4 }-6x^{ 2 }+9)$

$3x{ ({ x }^{ 2 }-3) }^{ 2 }$

$2{ x }^{ 3 }-50x$

$2x({ x }^{ 2 }-25)$

$2x({ x }^{ 2 }-{ 5 }^{ 2 })$

$2x[(x-5)(x+5)]$

$2{ x }^{ 5 }-32x$

$2x({ x }^{ 4 }-16)$

$2x[{ ({ x }^{ 2 }) }^{ 2 }-{ 4 }^{ 2 }]$

$2x[({ x }^{ 2 }-4)({ x }^{ 2 }+4)]$

$2x[({ x }^{ 2 }-{ 2 }^{ 2 })({ x }^{ 2 }+4)]$

$2x[({ x }-{ 2 })({ x }+{ 2 })({ x }^{ 2 }+4)]$

$2{ x }^{ 2 }+x-28=0$

$a{ x }^{ 2 }+bx+c=0$

$a=2,\quad b=1,\quad c=-28$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(1)+\sqrt { { (1 })^{ 2 }-4(2)(-28) } }{ 2(2) } ,\frac { -(1)-\sqrt { { { (1 }) }^{ 2 }-4(2)(-28) } }{ 2(2) }$

$x=\quad \frac { -1+\sqrt { { 1 }+224 } }{ 4 } ,\frac { -1-\sqrt { { 1 }+224 } }{ 4 }$

$x=\quad \frac { -1+\sqrt { 225 } }{ 4 } ,\frac { -1-\sqrt { 225 } }{ 4 }$

$x=\quad \frac { -1+15 }{ 4 } ,\frac { -1-15 }{ 4 }$

$x=\quad \frac { 14 }{ 4 } ,\frac { -16 }{ 4 }$

$x=\frac { 7 }{ 2 } \quad x=-4$

$2{ x }^{ 2 }+x-28=2(x+4)(x-\frac { 7 }{ 2 } )$

Solution of exercise 15

$\frac { 2 }{ 5 } { x }^{ 5 }-\frac { 6 }{ 5 } { x }^{ 4 }+\frac { 14 }{ 15 } { x }^{ 2 }$

$\frac { 2 }{ 5 } { x }^{ 2 }({ x }^{ 3 }-3{ x }^{ 2 }+\frac { 7 }{ 3 } )$

$2xy-4x-3y+6$

$2x(y-2)-3(y-2)$

$(2x-3)(y-2)$

$25{ x }^{ 2 }-1$

${ (5x) }^{ 2 }-{ 1 }^{ 2 }$

$(5x-1)(5x+1)$

$36{ x }^{ 2 }-49$

${ (6x) }^{ 2 }-7^{ 2 }$

$(6x-7)(6x+7)$

$5{ x }^{ 2 }-10x+5$

$5({ x }^{ 2 }-2x+1)$

$5[{ (x) }^{ 2 }-2(x)(1)+{ 1 }^{ 2 }]$

$5{ (x-1) }^{ 2 }$

${ x }^{ 2 }-6x+9$

${ x }^{ 2 }-2(x)(3)+{ 3 }^{ 2 }$

${ (x-3) }^{ 2 }$

${ x }^{ 2 }-20x+100$

${ x }^{ 2 }-2(x)(10)+{ 10 }^{ 2 }$

${ (x-10) }^{ 2 }$

${ x }^{ 2 }+10x+25$

${ x }^{ 2 }+2(x)(5)+{ 5 }^{ 2 }$

${ (x+5) }^{ 2 }$

${ x }^{ 2 }+14x+49$

${ x }^{ 2 }+2(x)(7)+{ 7 }^{ 2 }$

${ (x+7) }^{ 2 }$

${ x }^{ 3 }-4{ x }^{ 2 }+4x$

$x({ x }^{ 2 }-4x+4)$

$x({ x }^{ 2 }-2(x)(2)+{ 2 }^{ 2 })$

$x{ (x-2) }^{ 2 }$

$3{ x }^{ 7 }-27x$

$3x({ x }^{ 6 }-9)$

$3x[{ ({ x }^{ 3 }) }^{ 2 }-{ 3 }^{ 2 }]$

$3x{ ({ x }^{ 3 }-3) }({ x }^{ 3 }+3)$

${ x }^{ 2 }-11x+30=0$

$a{ x }^{ 2 }+bx+c=0$

$a=1,\quad b=-11,\quad c=30$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-11)+\sqrt { { (-11 })^{ 2 }-4(1)(30) } }{ 2(1) } ,\frac { -(-11)-\sqrt { { { (-11 }) }^{ 2 }-4(2)(30) } }{ 2(1) }$

$x=\quad \frac { 11+\sqrt { { 121 }-120 } }{ 2 } ,\frac { 11-\sqrt { { 121 }-120 } }{ 2 }$

$x=\quad \frac { 11+\sqrt { 1 } }{ 2 } ,\frac { 11-\sqrt { 1 } }{ 2 }$

$x=\quad \frac { 11+1 }{ 2 } ,\frac { 11-1 }{ 2 }$

$x=\quad \frac { 12 }{ 2 } ,\frac { 10 }{ 2 }$

$x=6\quad x=5$

${ x }^{ 2 }-11x+30=(x-5)(x-6)$

$3{ x }^{ 2 }-10x+3=0$

$a{ x }^{ 2 }+bx+c=0$

$a=3,\quad b=-10,\quad c=3$

$x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\quad \frac { -(-10)+\sqrt { { (-10 })^{ 2 }-4(3)(3) } }{ 2(3) } ,\frac { -(-10)-\sqrt { { { (-10 }) }^{ 2 }-4(3)(3) } }{ 2(3) }$

$x=\quad \frac { 10+\sqrt { { 100 }-36 } }{ 6 } ,\frac { 10-\sqrt { { 100 }-36 } }{ 6 }$

$x=\quad \frac { 10+\sqrt { 64 } }{ 6 } ,\frac { 10-\sqrt { 64 } }{ 6 }$

$x=\quad \frac { 10+8 }{ 6 } ,\frac { 10-8 }{ 6 }$

$x=\quad \frac { 18 }{ 6 } ,\frac { 2 }{ 6 }$

$x=3\quad x=\frac { 1 }{ 3 }$

$3{ x }^{ 2 }-10x+3=(x-3)(x-\frac { 1 }{ 3 } )$

$2{ x }^{ 2 }-x-1=0$

$a{ x }^{ 2 }+bx+c=0$

$a=2,\quad b=-1,\quad c=-1$

$x=\quad \frac { -(-1)+\sqrt { { (-1 })^{ 2 }-4(2)(-1) } }{ 2(2) } ,\frac { -(-1)-\sqrt { { { (-1 }) }^{ 2 }-4(2)(-1) } }{ 2(2) }$

$x=\quad \frac { 1+\sqrt { { 1 }+8 } }{ 4 } ,\frac { 1-\sqrt { { 1 }+8 } }{ 4 }$

$x=\quad \frac { 1+\sqrt { 9 } }{ 4 } ,\frac { 1-\sqrt { 9 } }{ 4 }$

$x=\quad \frac { 1+3 }{ 4 } ,\frac { 1-3 }{ 4 }$

$x=\quad \frac { 4 }{ 4 } ,\frac { -2 }{ 4 }$

$x=1\quad x=-\frac { 1 }{ 2 }$

$2{ x }^{ 2 }-x-1=(x-1)(x+\frac { 1 }{ 2 } )$

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Factoring Polynomials Worksheet

Factor and Calculate the Roots of the Following Polynomials

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10

Exercise 11

Exercise 12

Exercise 13

Exercise 14

Exercise 15

Solution of exercise 1

Solution of exercise 2

Solution of exercise 3

Solution of exercise 4

Solution of exercise 5

Solution of exercise 6

Solution of exercise 7

Solution of exercise 8

Solution of exercise 9

Solution of exercise 10

Solution of exercise 11

Solution of exercise 12

Solution of exercise 13

Solution of exercise 14

Solution of exercise 15

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